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Ppy3N7ZDNZ9Ix0Av | maths | complex-numbers | nth-roots-of-unity | The value of $$\sum\limits_{k = 1}^{10} {\left( {\sin {{2k\pi } \over {11}} + i\,\,\cos {{2k\pi } \over {11}}} \right)} $$ is : | [{"identifier": "A", "content": "i"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "- 1"}, {"identifier": "D", "content": "- i"}] | ["D"] | null | $$\sum\limits_{k = 1}^{10} {\left( {\sin {{2k\pi } \over {11}} + i\cos {{2k\pi } \over {11}}} \right)} $$
<br><br>$$ = i\sum\limits_{k = 1}^{10} {\left( {\cos {{2k\pi } \over {11}} - i\,\sin {{2k\pi } \over {11}}} \right)} $$
<br><br>$$ = i\sum\limits_{k = 1}^{10} {{e^{ - {{2k\pi } \over {11}}}}} i = i\left\{ {\sum\limits_{k = 0}^{10} {{e^{ - {{2k\pi } \over {11}}}}} - 1} \right\}$$
<br><br>$$ = i\left[ {1 + {e^{ - {{2\pi } \over {11}}i}} + e - {{4\pi } \over {11}}i + .....11\,\,terms} \right] - i$$
<br><br>$$ = i\left[ {{{1 - {{\left( {{e^{ - {{2\pi } \over {11}}}}} \right)}^{11}}} \over {1 - {e^{ - {{2\pi } \over {11}}i}}}}} \right] - i$$
<br><br>$$ = i\left[ {{{1 - {e^{ - 2\pi i}}} \over {1 - {e^{ - {{2\pi } \over {11}}i}}}}} \right] - i$$
<br><br>$$ = i \times 0 - i$$
<br><br>[as $$\,\,\,\,\,\,$$ $${e^{ - 2\pi i}} = 1$$ ]
<br><br>$$ = - i$$ | mcq | aieee-2006 |
1dXMQ34Gy45CNn7DdBZAT | maths | complex-numbers | nth-roots-of-unity | Let $$\alpha $$ and $$\beta $$ be two roots of the equation x<sup>2</sup> + 2x + 2 = 0 , then $$\alpha ^{15}$$ + $$\beta ^{15}$$ is equal to : | [{"identifier": "A", "content": "-256"}, {"identifier": "B", "content": "512"}, {"identifier": "C", "content": "-512"}, {"identifier": "D", "content": "256"}] | ["A"] | null | Given equation,
<br><br>x<sup>2</sup> + 2x + 2 = 0
<br><br>$$ \therefore $$ x = $${{ - 2 \pm \sqrt {4 - 4.1.2} } \over {2.1}}$$
<br><br>x = $$-$$ 1 $$ \pm $$ i
<br><br>$$ \therefore $$ $$\alpha $$ = $$-$$ 1 + i
<br><br>and $$\beta $$ = $$-$$ 1 $$-$$ i
<br><br><u>Note </u> :
<br><br>x + iy = r (cos$$\theta $$ + isin$$\theta $$)
<br><br>$$ \therefore $$ (x + iy)<sup>n</sup> = r<sup>n</sup> (cosn$$\theta $$ + isinn$$\theta $$)
<br><br>$$ \therefore $$ $$-$$ 1 + i = $$\sqrt 2 $$ [cos$${{3\pi } \over 4}$$ + isin$${{3\pi } \over 4}$$ ]
<br><br>$$ \Rightarrow $$ ($$-$$ 1 + i)<sup>15</sup> = $${\left( {\sqrt 2 } \right)^{15}}$$ [cos$$\left( {{{15.3\pi } \over 4}} \right) + i\sin \left( {{{15.3\pi } \over 4}} \right)$$]
<br><br>And $$-$$1 $$-$$ i = $$\sqrt 2 $$ $$\left[ {\cos \left( { - {{3\pi } \over 4}} \right) + i\sin \left( { - {{3\pi } \over 4}} \right)} \right]$$
<br><br>= $$\sqrt 2 \left[ {\cos {{3\pi } \over 4} - \sin {{3\pi } \over 4}} \right]$$
<br><br>$$ \therefore $$ ($$-$$1 $$-$$ i)<sup>15</sup> = $${\left( {\sqrt 2 } \right)^{15}}\left[ {\cos \left( {{{15.3\pi } \over 4}} \right) - i\sin \left( {{{15.3\pi } \over 4}} \right)} \right]$$
<br><br>Now
<br><br>$$\alpha $$<sup>15</sup> + $$\beta $$<sup>15</sup>
<br><br>= ($$-$$1 + i)<sup>15</sup> + ($$-$$ 1 $$-$$ i)<sup>15</sup>
<br><br>= $${\left( {\sqrt 2 } \right)^{15}}$$ $$\left[ {2\cos \left( {{{15.3\pi } \over 4}} \right)} \right]$$
<br><br>= $${\left( {\sqrt 2 } \right)^{15}}\left[ {2\cos \left( {11\pi + {\pi \over 4}} \right)} \right]$$
<br><br>= $${\left( {\sqrt 2 } \right)^{15}}\left[ {2\left( { - \cos {\pi \over 4}} \right)} \right]$$
<br><br>= $${\left( {\sqrt 2 } \right)^{15}} \times 2 \times - {1 \over {\sqrt 2 }}$$
<br><br>= $$ - {\left( {\sqrt 2 } \right)^{14}}.2$$
<br><br>= $$-$$ 2<sup>7</sup> $$ \times $$ 2
<br><br>= $$-$$ 2<sup>8</sup>
<br><br>= $$-$$ 256 | mcq | jee-main-2019-online-9th-january-morning-slot |
noAYQK9TnN7SiKwnAIFy2 | maths | complex-numbers | nth-roots-of-unity | If $$\alpha $$ and $$\beta $$ be the roots of the equation
x<sup>2</sup> – 2x + 2 = 0, then the least value of n for which $${\left( {{\alpha \over \beta }} \right)^n} = 1$$ is :
| [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "3"}] | ["C"] | null | x<sup>2</sup> – 2x + 2 = 0
<br><br>$$ \therefore $$ x = $${{2 \pm \sqrt { - 4} } \over 2} = 1 \pm i$$
<br><br>Now, $${\alpha \over \beta } = {{1 + i} \over {1 - i}} = {{{{\left( {1 + i} \right)}^2}} \over {1 - {i^2}}} = i$$
<br><br>or $${\alpha \over \beta } = {{1 - i} \over {1 + i}} = {{{{\left( {1 - i} \right)}^2}} \over {1 + {i^2}}} = - i$$
<br><br>$$ \therefore $$ $${\left( { i} \right)^n}$$ = 1
<br><br>and $${\left( { - i} \right)^n}$$ = 1
<br><br>So n must be a multiple of 4
<br><br>$$ \therefore $$ Minimum value of n = 4
| mcq | jee-main-2019-online-8th-april-morning-slot |
DLDIcZtF5DiyreYTGnjgy2xukf0zyu08 | maths | complex-numbers | nth-roots-of-unity | If $${\left( {{{1 + i} \over {1 - i}}} \right)^{{m \over 2}}} = {\left( {{{1 + i} \over {1 - i}}} \right)^{{n \over 3}}} = 1$$, (m, n
$$ \in $$ N) then the
greatest common divisor of the least values of
m and n is _______ . | [] | null | 4 | $${\left( {{{1 + i} \over {1 - i}}} \right)^{m/2}} = {\left( {{{1 + i} \over {1 - i}}} \right)^{n/3}} = 1$$<br><br>$$ \Rightarrow {\left( {{{{{\left( {1 + i} \right)}^2}} \over 2}} \right)^{m/2}} = {\left( {{{{{\left( {1 + i} \right)}^2}} \over { - 2}}} \right)^{n/3}} = 1$$<br><br>$$ \Rightarrow {(i)^{m/2}} = {( - i)^{n/3}} = 1$$ = i<sup>4</sup> [ As i<sup>4</sup> = 1]<br><br>$$ \Rightarrow {m \over 2} = 4{k_1}\,and\,{n \over 3} = 4{k_2}$$<br><br>$$ \Rightarrow $$ m = 8 k<sub>1</sub> and n = 12 k<sub>2</sub><br><br>Least value of m = 8 and n = 12.<br><br>$$ \therefore $$ GCD (8, 12) = 4 | integer | jee-main-2020-online-3rd-september-morning-slot |
1ktbgurkn | maths | complex-numbers | nth-roots-of-unity | Let $$z = {{1 - i\sqrt 3 } \over 2}$$, $$i = \sqrt { - 1} $$. Then the value of $$21 + {\left( {z + {1 \over z}} \right)^3} + {\left( {{z^2} + {1 \over {{z^2}}}} \right)^3} + {\left( {{z^3} + {1 \over {{z^3}}}} \right)^3} + .... + {\left( {{z^{21}} + {1 \over {{z^{21}}}}} \right)^3}$$ is ______________. | [] | null | 13 | $$z = {{1 - \sqrt {3i} } \over 2} = {e^{ - i{\pi \over 3}}}$$<br><br>$${z^r} + {1 \over {{z^r}}} = 2\cos \left( { - {\pi \over 3}} \right)r = 2\cos {{r\pi } \over 3}$$<br><br>$$ \Rightarrow 21 + \sum\limits_{r = 1}^{21} {{{\left( {{z^r} + {1 \over {{z^r}}}} \right)}^3} = 8\left( {{{\cos }^3}{{r\pi } \over 3}} \right) = 2\left( {\cos r\pi + 3\cos {{r\pi } \over 3}} \right)} $$<br><br>$$ \Rightarrow 21 + {\left( {z + {1 \over 2}} \right)^3} + {\left( {{z^2} + {1 \over {{z^2}}}} \right)^3} + ....{\left( {{z^{21}} + {1 \over {{z^{21}}}}} \right)^3}$$<br><br>$$ = 21 + \sum\limits_{r = 1}^{21} {{{\left( {{z^r} + {1 \over {{z^r}}}} \right)}^3}} $$<br><br>$$ = 21 + \sum\limits_{r = 1}^{21} {\left( {2\cos r\pi + 6\cos {{r\pi } \over 3}} \right)} $$<br><br>$$ = 21 - 2 - 6$$<br><br>$$ = 13$$ | integer | jee-main-2021-online-26th-august-morning-shift |
1lsg3wb0w | maths | complex-numbers | nth-roots-of-unity | <p>If $$z$$ is a complex number, then the number of common roots of the equations $$z^{1985}+z^{100}+1=0$$ and $$z^3+2 z^2+2 z+1=0$$, is equal to</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "3"}] | ["B"] | null | <p>$$\begin{array}{ll}
\text { } & z^{1985}+z^{100}+1=0 \& z^3+2 z^2+2 z+1=0 \\
& (z+1)\left(z^2-z+1\right)+2 z(z+1)=0 \\
& (z+1)\left(z^2+z+1\right)=0 \\
\Rightarrow \quad & z=-1, \quad z=w, w^2 \\
& \text { Now putting } z=-1 \text { not satisfy } \\
& \text { Now put } z=w \\
\Rightarrow \quad & w^{1985}+w^{100}+1 \\
\Rightarrow \quad & w^2+w+1=0 \\
\Rightarrow \quad & \text { Also, } z=w^2 \\
\Rightarrow \quad & w^{3970}+w^{200}+1 \\
\Rightarrow & w+w^2+1=0
\end{array}$$</p>
<p>Two common root</p> | mcq | jee-main-2024-online-30th-january-evening-shift |
lv5gsy4m | maths | complex-numbers | nth-roots-of-unity | <p>If the set $$R=\{(a, b): a+5 b=42, a, b \in \mathbb{N}\}$$ has $$m$$ elements and $$\sum_\limits{n=1}^m\left(1-i^{n !}\right)=x+i y$$, where $$i=\sqrt{-1}$$, then the value of $$m+x+y$$ is</p> | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "5"}] | ["A"] | null | <p>$$R=\{(a, b): a+5 b=42\}$$</p>
<p>Then $$R=\{(2,8),(7,7),(12,6),(17,5),(22,4),(27, 3),(32,2),(37,1)\}$$</p>
<p>$$\begin{aligned}
& \text { and } \sum_{n=1}^{\substack{m=8}}\left(1-i^{n!}\right)=x+i y \\
& \therefore \sum_{n=1}^8\left(1-i^{n!}\right)=8-\left(i+i^2+i^6+1+1+1+1+1\right) \\
& =5-i \\
& \therefore x=5, y=-1 \\
& x+y+m=5-1+8=12
\end{aligned}$$</p> | mcq | jee-main-2024-online-8th-april-morning-shift |
jO4bSQV4DP64o4Ji8Ljgy2xukezaaujn | maths | complex-numbers | square-root-of-a-complex-number | The imaginary part of
<br/>$${\left( {3 + 2\sqrt { - 54} } \right)^{{1 \over 2}}} - {\left( {3 - 2\sqrt { - 54} } \right)^{{1 \over 2}}}$$ can be : | [{"identifier": "A", "content": "-2$$\\sqrt 6 $$"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "$$\\sqrt 6 $$"}, {"identifier": "D", "content": "-$$\\sqrt 6 $$"}] | ["A"] | null | $$3 + 2\sqrt { - 54} $$<br><br>
$$ = 9 - 6 + 2\sqrt { - 54} $$<br><br>
$$ = 9 + {\left( {\sqrt 6 i} \right)^2} + 2.3.\sqrt 6 i$$<br><br>
$$ = {3^2} + {\left( {\sqrt 6 i} \right)^2} + 2.3.\sqrt 6 i$$<br><br>
$$ = {\left( {3 + \sqrt 6 i} \right)^2}$$<br><br>
Similarly, $$\left( {3 - 2\sqrt { - 54} } \right) = {\left( {3 - \sqrt 6 i} \right)^2}$$<br><br>
$$ \therefore {\left( {3 + 2\sqrt { - 54} } \right)^{{1 \over 2}}} - {\left( {3 - 2\sqrt { - 54} } \right)^{{1 \over 2}}}$$<br><br>
$$ = \pm \left( {3 + \sqrt 6 i} \right) - \left[ { \pm \left( {3 - \sqrt 6 i} \right)} \right]$$<br><br>
$$ = 6, - 6,2\sqrt 6 i, - 2\sqrt 6 i$$<br><br>
$$ \therefore $$ Possible imaginary parts are $$2\sqrt 6 i, - 2\sqrt 6 i$$ | mcq | jee-main-2020-online-2nd-september-evening-slot |
JQ9MUck61xQ6DdPr | maths | definite-integration | definite-integral-as-a-limit-of-sum | $$\mathop {\lim }\limits_{n \to \infty } {{{1^p} + {2^p} + {3^p} + ..... + {n^p}} \over {{n^{p + 1}}}}$$ is | [{"identifier": "A", "content": "$${1 \\over {p + 1}}$$"}, {"identifier": "B", "content": "$${1 \\over {1 - p}}$$"}, {"identifier": "C", "content": "$${1 \\over p} - {1 \\over {p - 1}}$$"}, {"identifier": "D", "content": "$${1 \\over {p + 2}}$$"}] | ["A"] | null | We have $$\mathop {\lim }\limits_{x \to \infty } {{{1^p} + {2^p} + .... + {n^p}} \over {{n^{p + 1}}}};$$
<br><br>$$\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^n {{{{r^p}} \over {{n^p}.n}}} = \int\limits_0^1 {{x^p}} dx$$
<br><br>$$ = {\left[ {{{{x^{p + 1}}} \over {p + 1}}} \right]_0} = {1 \over {p + 1}}$$ | mcq | aieee-2002 |
wXfQ6ecjBBTNFejK | maths | definite-integration | definite-integral-as-a-limit-of-sum | $$\mathop {\lim }\limits_{n \to \infty } {{1 + {2^4} + {3^4} + .... + {n^4}} \over {{n^5}}}$$ - $$\mathop {\lim }\limits_{n \to \infty } {{1 + {2^3} + {3^3} + .... + {n^3}} \over {{n^5}}}$$ | [{"identifier": "A", "content": "$${1 \\over 5}$$"}, {"identifier": "B", "content": "$${1 \\over 30}$$"}, {"identifier": "C", "content": "zero"}, {"identifier": "D", "content": "$${1 \\over 4}$$"}] | ["A"] | null | The given expression can be written as
<br><br>$$\mathop {\lim }\limits_{n \to \infty } {1 \over n}{\sum\limits_{r = 1}^n {\left( {{r \over n}} \right)} ^4} - \mathop {\lim }\limits_{n \to \infty } {1 \over n}.\mathop {\lim }\limits_{n \to \infty } {1 \over n}{\left( {{r \over n}} \right)^3}$$
<br><br>$$ = \int\limits_0^1 {{x^4}} \,\,dx - \mathop {\lim }\limits_{n \to \infty } {1 \over n} \times \int\limits_0^1 {{x^3}} \,\,dx$$
<br><br>$$ = \left[ {{{{x^5}} \over 5}} \right]_0^1 - 0 = {1 \over 5}$$ | mcq | aieee-2003 |
ze5mgsSsHsysqEx4 | maths | definite-integration | definite-integral-as-a-limit-of-sum | $$\mathop {Lim}\limits_{n \to \infty } \sum\limits_{r = 1}^n {{1 \over n}{e^{{r \over n}}}} $$ is | [{"identifier": "A", "content": "$$e+1$$ "}, {"identifier": "B", "content": "$$e-1$$ "}, {"identifier": "C", "content": "$$1-e$$"}, {"identifier": "D", "content": "$$e$$"}] | ["B"] | null | $$\mathop {Lim}\limits_{n \to \infty } \sum\limits_{r = 1}^n {{1 \over n}} {e^{{r \over n}}}\,\,$$
<br><br>$$\left[ {} \right.$$ Using definite integrals as limit of sum $$\left. {} \right]$$
<br><br>$$ = \int\limits_\theta ^1 {{e^x}} dx = e - 1$$ | mcq | aieee-2004 |
zQuI65ExuF62vE2u | maths | definite-integration | definite-integral-as-a-limit-of-sum | $$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over {{n^2}}}{{\sec }^2}{1 \over {{n^2}}} + {2 \over {{n^2}}}{{\sec }^2}{4 \over {{n^2}}}.... + {1 \over n}{{\sec }^2}1} \right]$$
<br/>equals | [{"identifier": "A", "content": "$${1 \\over 2}\\sec 1$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$cosec 1"}, {"identifier": "C", "content": "tan 1"}, {"identifier": "D", "content": "$${1 \\over 2}$$tan 1"}] | ["D"] | null | $$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over {{n^2}}}{{\sec }^2}{1 \over {{n^2}}} + {2 \over {{n^2}}}{{\sec }^2}{4 \over {{n^2}}} + {3 \over {{n^2}}}se{c^2}{9 \over {{n^2}}} + ... + {1 \over n}{{\sec }^2}1} \right]$$
<br><br>$$ = \mathop {\lim }\limits_{n \to \infty } {r \over {{n^2}}}{\sec ^2}{{{r^2}} \over {{n^2}}} = \mathop {\lim }\limits_{n \to \infty } {1 \over n}.{\pi \over n}{\sec ^2}{{{r^2}} \over {{n^2}}}$$
<br><br>$$ \Rightarrow $$ Given limit is equal to value of integral
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int\limits_0^1 {x{{\sec }^2}} \,{x^2}\,dx$$
<br><br>or $${1 \over 2}\int\limits_0^1 {2x\,\sec } \,\,{x^2}dx = {1 \over 2}\int\limits_0^1 {{{\sec }^2}} tdl$$
<br><br>$$\left[ {} \right.$$ put $$\left. {{x^2} = t} \right]$$
<br><br>$$ = {1 \over 2}\left( {\tan t} \right)_0^1 = {1 \over 2}\tan \,1.$$ | mcq | aieee-2005 |
VgQrb1CpLwDux43W | maths | definite-integration | definite-integral-as-a-limit-of-sum | $$\mathop {\lim }\limits_{n \to \infty } {\left( {{{\left( {n + 1} \right)\left( {n + 2} \right)...3n} \over {{n^{2n}}}}} \right)^{{1 \over n}}}$$ is equal to: | [{"identifier": "A", "content": "$${9 \\over {{e^2}}}$$ "}, {"identifier": "B", "content": "$$3\\,\\log \\,3 - 2$$ "}, {"identifier": "C", "content": "$${{18} \\over {{e^4}}}$$ "}, {"identifier": "D", "content": "$${{27} \\over {{e^2}}}$$ "}] | ["D"] | null | $$y = \mathop {\lim }\limits_{n \to \infty } {\left( {{{\left( {n + 1} \right)\left( {n + 2} \right)...3n} \over {{n^{2n}}}}} \right)^{{1 \over n {}}}}$$
<br><br>$$\ln \,y = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\ln \left( {1 + {1 \over n}} \right)\left( {1 + {2 \over n}} \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....\left( {1 + {2 \over n}} \right)$$
<br><br>$$\ln {\mkern 1mu} y = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\left[ {\ln \left( {1 + {1 \over n}} \right) + \ln \left( {1 + {2 \over n}} \right)} \right.$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. { + .... + \ln \left( {1 + {{2n} \over n}} \right)} \right]$$
<br><br>$$ = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 1}^{2n} {\ln \left( {1 + {r \over n}} \right)} $$
<br><br>$$ = \int_0^2 {\ln \left( {1 + x} \right)dx} $$
<br><br>Let $$1 + x = t \Rightarrow dx = dt$$
<br><br>when $$x = 0,t = 1$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,x = 2,\,\,t = 3$$
<br><br>$$\ln \,y = \int_1^3 {\ln t\,d\,t = } \left[ {t\,\ln \,t - \left. {t_1^3} \right]} \right.$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\, = \ln \left( {{{{3^3}} \over {{e^2}}}} \right) = \ln \left( {{{27} \over {{e^2}}}} \right)$$
<br><br>$$ \Rightarrow y = {{27} \over {{e^2}}}$$ | mcq | jee-main-2016-offline |
UerCkxqwxx08Smvfk0RMh | maths | definite-integration | definite-integral-as-a-limit-of-sum | If $$\mathop {\lim }\limits_{n \to \infty } \,\,{{{1^a} + {2^a} + ...... + {n^a}} \over {{{(n + 1)}^{a - 1}}\left[ {\left( {na + 1} \right) + \left( {na + 2} \right) + ..... + \left( {na + n} \right)} \right]}} = {1 \over {60}}$$
<br/><br/>for some positive real number a, then a is equal to :
| [{"identifier": "A", "content": "7 "}, {"identifier": "B", "content": "8 "}, {"identifier": "C", "content": "$${{15} \\over 2}$$"}, {"identifier": "D", "content": "$${{17} \\over 2}$$"}] | ["A"] | null | $$\mathop {\lim }\limits_{n \to \infty } {{{1 \over {\left( {a + 1} \right)}}\,.\,{n^{a + 1}} + {a_1}{n^a} + {a_2}{n^{a - 1}} + .\,.\,.\,.} \over {{{\left( {n + 1} \right)}^{a - 1}}.\,{n^2}\left( {a + {{1 + {1 \over n}} \over 2}} \right)}} = {1 \over {60}}$$
<br><br>$$ \Rightarrow $$ $$\mathop {\lim }\limits_{n \to \infty } {{{{\left( {{1 \over n}} \right)}^2} + {{\left( {{2 \over n}} \right)}^a} + ....... + {{\left( {{n \over n}} \right)}^a}} \over {{{\left( {n + 1} \right)}^{a - 1}}\left[ {{a^2}a + {{n\left( {n + 1} \right)} \over 2}} \right]}} = {1 \over {60}}$$
<br><br> $$ = {{\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 1}^n {{{\left( {{r \over n}} \right)}^a}} } \over {{{\left( {1 + {1 \over n}} \right)}^{a - 1}}\left[ {a + {1 \over 2}\left( {1 + {1 \over n}} \right)} \right]}} = {1 \over {60}}$$
<br><br> $$ = {{\int_0^1 {{x^a}dx} } \over {\left( {a + {1 \over 2}} \right)}} = {1 \over {60}} = {{{1 \over {a + 1}}} \over {a + {1 \over 2}}} = {1 \over {60}}$$
<br><br>$$ \Rightarrow $$ $${{{1 \over {a + 1}}} \over {\left( {a + {1 \over 2}} \right)}} = {1 \over {60}} \Rightarrow \left( {a + 1} \right)\left( {2a + 1} \right) = 120$$
<br><br>$$ \Rightarrow $$ 2a<sup>2</sup> + 3a $$-$$ 119 $$=$$ 0
<br><br>$$ \Rightarrow $$ 2a<sup>2</sup> + 17a $$-$$ 14a $$-$$ 119 $$=$$ 0
<br><br>$$ \Rightarrow $$ (a $$-$$ 7) (2a + 17) $$=$$ 0
<br><br.$$ \rightarrow="" $$ a="" $$="$$" 7,="" $$-$$="" $${{17}="" \over="" 2}$$="" <br=""><br>$$ \Rightarrow $$ a $$=$$ 7, $$-$$ $${{17} \over 2}$$</br.$$> | mcq | jee-main-2017-online-9th-april-morning-slot |
Dk9DSWeqwOHhPHwxzNN0b | maths | definite-integration | definite-integral-as-a-limit-of-sum | $$\mathop {\lim }\limits_{x \to \infty } \left( {{n \over {{n^2} + {1^2}}} + {n \over {{n^2} + {2^2}}} + {n \over {{n^2} + {3^2}}} + ..... + {1 \over {5n}}} \right)$$ is equal to : | [{"identifier": "A", "content": "tan<sup>\u20131 </sup>(2)"}, {"identifier": "B", "content": "tan<sup>\u20131 </sup>(3)"}, {"identifier": "C", "content": "$${\\pi \\over 4}$$"}, {"identifier": "D", "content": "$${\\pi \\over 2}$$"}] | ["A"] | null | $$\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^{2n} {{n \over {{n^2} + {r^2}}}} $$
<br><br>$$\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^{2n} {{1 \over {n\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}} = \int\limits_0^2 {{{dx} \over {1 + {x^2}}}} } = {\tan ^{ - 1}}2$$ | mcq | jee-main-2019-online-12th-january-evening-slot |
hSHUIycDWkH38RPecf3rsa0w2w9jwy2hm13 | maths | definite-integration | definite-integral-as-a-limit-of-sum | $$\mathop {\lim }\limits_{n \to \infty } \left( {{{{{(n + 1)}^{1/3}}} \over {{n^{4/3}}}} + {{{{(n + 2)}^{1/3}}} \over {{n^{4/3}}}} + ....... + {{{{(2n)}^{1/3}}} \over {{n^{4/3}}}}} \right)$$
<br/>is equal to : | [{"identifier": "A", "content": "$${4 \\over 3}{\\left( 2 \\right)^{3/4}}$$"}, {"identifier": "B", "content": "$${3 \\over 4}{\\left( 2 \\right)^{4/3}} - {3 \\over 4}$$"}, {"identifier": "C", "content": "$${4 \\over 3}{\\left( 2 \\right)^{4/3}}$$"}, {"identifier": "D", "content": "$${3 \\over 4}{\\left( 2 \\right)^{4/3}} - {4 \\over 3}$$"}] | ["B"] | null | $$\mathop {\lim }\limits_{n \to \infty } {{{{(n + 1)}^{{1 \over 3}}} + {{(n + 2)}^{{1 \over 3}}} + ..... + {{\left( {n + n} \right)}^{{1 \over 3}}}} \over {n{{(n)}^{{1 \over 3}}}}}$$<br><br>
$$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{{{{(n + r)}^{{1 \over 3}}}} \over {n.{n^{{1 \over 3}}}}}\,\,\,{r \over n} \to x\,and\,{1 \over n} \to dx\,} $$<br><br>
$$ \Rightarrow \int\limits_0^1 {{{(1 + x)}^{{1 \over 3}}}dx} $$<br><br>
$$ \Rightarrow \left[ {{3 \over 4}{{(1 + x)}^{{4 \over 3}}}} \right]_0^1 = {3 \over 4}{(2)^{{4 \over 3}}} - {3 \over 4}$$ | mcq | jee-main-2019-online-10th-april-morning-slot |
KPNhsqLgDWKQZvRCQ31klt9lykx | maths | definite-integration | definite-integral-as-a-limit-of-sum | $$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over n} + {n \over {{{(n + 1)}^2}}} + {n \over {{{(n + 2)}^2}}} + ........ + {n \over {{{(2n + 1)}^2}}}} \right]$$ is equal to : | [{"identifier": "A", "content": "$${{1 \\over 2}}$$"}, {"identifier": "B", "content": "$${{1 \\over 3}}$$"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "$${{1 \\over 4}}$$"}] | ["A"] | null | $$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over n} + {n \over {{{(n + 1)}^2}}} + {n \over {{{(n + 2)}^2}}} + ... + {n \over {{{(2n - 1)}^2}}}} \right]$$
<br><br>$$\mathop {\lim }\limits_{n \to \infty } \left[ {{n \over {{{\left( {n + 0} \right)}^2}}} + {n \over {{{\left( {n + 1} \right)}^2}}} + {n \over {{{\left( {n + 2} \right)}^2}}} + ... + {n \over {{{\left( {n + \left( {n - 1} \right)} \right)}^2}}}} \right]$$
<br><br>$$ = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {{n \over {{{(n + r)}^2}}}} = $$$$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {{n \over {{n^2}{{\left( {1 + {r \over n}} \right)}^2}}}} $$<br><br>$$ = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{n - 1} {{1 \over {{{(r/n)}^2} + 2(r/n) + 1}}} $$<br><br>$$ = \int\limits_0^1 {{{dx} \over {{{(x + 1)}^2}}} = \left[ {{{ - 1} \over {(x + 1)}}} \right]_0^1 = {1 \over 2}} $$
<br><br><b>Note :</b> $$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^{pn} {{1 \over n}f\left( {{r \over n}} \right)} = \int_\alpha ^\beta {f(x)} dx$$<br><br>where, $$\alpha = \mathop {\lim }\limits_{n \to \infty } {r \over n} = 0$$ (as r = 1)<br><br>and $$\beta = \mathop {\lim }\limits_{n \to \infty } {r \over n} = p$$ (as r = pn)
<br><br>Here $$\alpha = \mathop {\lim }\limits_{n \to \infty } {r \over n} = 0$$ (as r = 0)
<br><br>and $$\beta = \mathop {\lim }\limits_{n \to \infty } {r \over n} = \mathop {\lim }\limits_{n \to \infty } {{n - 1} \over n}$$ = 1 (as r = n - 1) | mcq | jee-main-2021-online-25th-february-evening-slot |
gaG80xeroFIPmrjcDJ1kmhzhgz7 | maths | definite-integration | definite-integral-as-a-limit-of-sum | Let f : (0, 2) $$ \to $$ R be defined as f(x) = log<sub>2</sub>$$\left( {1 + \tan \left( {{{\pi x} \over 4}} \right)} \right)$$. Then, $$\mathop {\lim }\limits_{n \to \infty } {2 \over n}\left( {f\left( {{1 \over n}} \right) + f\left( {{2 \over n}} \right) + ... + f(1)} \right)$$ is equal to ___________. | [] | null | 1 | $$E = 2\mathop {\lim }\limits_{x \to \infty } \sum\limits_{r = 1}^n {{1 \over n}} f\left( {{r \over n}} \right)$$<br><br>$$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( {1 + \tan {{\pi x} \over 4}} \right)dx} $$ ..... (i)<br><br>replacing x $$ \to $$ 1 $$-$$ x<br><br>$$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( {1 + \tan {\pi \over 4}(1 - x)} \right)dx} $$<br><br>$$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( {1 + \tan \left( {{\pi \over 4} - {\pi \over 4}x} \right)} \right)dx} $$<br><br>$$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( {1 + {{1 - \tan {\pi \over 4}x} \over {1 + \tan {\pi \over 4}x}}} \right)dx} $$<br><br>$$E = {2 \over {\ln 2}}\int_0^1 {\ln \left( { {2 \over {1 + \tan {{\pi x} \over 4}}}} \right)dx} $$<br><br>$$E = {2 \over {\ln 2}}\int_0^1 {\left( {\ln 2 - \ln \left( {1 + \tan {{\pi x} \over 4}} \right)} \right)dx} $$ ..... (ii)<br><br>equation (i) + (ii)<br><br>2E = 2 $$ \Rightarrow $$ E = 1 | integer | jee-main-2021-online-16th-march-morning-shift |
1ks04wdm1 | maths | definite-integration | definite-integral-as-a-limit-of-sum | The value of $$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{j = 1}^n {{{(2j - 1) + 8n} \over {(2j - 1) + 4n}}} $$ is equal to : | [{"identifier": "A", "content": "$$5 + {\\log _e}\\left( {{3 \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$$2 - {\\log _e}\\left( {{2 \\over 3}} \\right)$$"}, {"identifier": "C", "content": "$$3 + 2{\\log _e}\\left( {{2 \\over 3}} \\right)$$"}, {"identifier": "D", "content": "$$1 + 2{\\log _e}\\left( {{3 \\over 2}} \\right)$$"}] | ["D"] | null | $$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{j = 1}^n {{{\left( {{{2j} \over n} - {1 \over n} + 8} \right)} \over {\left( {{{2j} \over n} - {1 \over n} + 4} \right)}}} $$<br><br>$$\int\limits_0^1 {{{2x + 8} \over {2x + 4}}dx = \int\limits_0^1 {dx + \int\limits_0^1 {{4 \over {2x + 4}}} dx} } $$<br><br>$$ = 1 + 4{1 \over 2}(\ln |2x + 4|)_0^1$$<br><br>$$ = 1 + 2\ln \left( {{3 \over 2}} \right)$$ | mcq | jee-main-2021-online-27th-july-morning-shift |
1ktbg2i67 | maths | definite-integration | definite-integral-as-a-limit-of-sum | The value of <br/><br/>$$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{2n - 1} {{{{n^2}} \over {{n^2} + 4{r^2}}}} $$ is : | [{"identifier": "A", "content": "$${1 \\over 2}{\\tan ^{ - 1}}(2)$$"}, {"identifier": "B", "content": "$${1 \\over 2}{\\tan ^{ - 1}}(4)$$"}, {"identifier": "C", "content": "$${\\tan ^{ - 1}}(4)$$"}, {"identifier": "D", "content": "$${1 \\over 4}{\\tan ^{ - 1}}(4)$$"}] | ["B"] | null | $$L = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{2n - 1} {{1 \over {1 + 4{{\left( {{r \over n}} \right)}^2}}}} $$<br><br>$$ \Rightarrow L = \int\limits_0^2 {{1 \over {1 + 4{x^2}}}dx} $$<br><br>$$ \Rightarrow L = \left. {{1 \over 2}{{\tan }^{ - 1}}(2x)} \right|_0^2 \Rightarrow L = {1 \over 2}{\tan ^{ - 1}}4$$ | mcq | jee-main-2021-online-26th-august-morning-shift |
1ktekp01k | maths | definite-integration | definite-integral-as-a-limit-of-sum | If $${U_n} = \left( {1 + {1 \over {{n^2}}}} \right)\left( {1 + {{{2^2}} \over {{n^2}}}} \right)^2.....\left( {1 + {{{n^2}} \over {{n^2}}}} \right)^n$$, then $$\mathop {\lim }\limits_{n \to \infty } {({U_n})^{{{ - 4} \over {{n^2}}}}}$$ is equal to : | [{"identifier": "A", "content": "$${{{e^2}} \\over {16}}$$"}, {"identifier": "B", "content": "$${4 \\over e}$$"}, {"identifier": "C", "content": "$${{16} \\over {{e^2}}}$$"}, {"identifier": "D", "content": "$${4 \\over {{e^2}}}$$"}] | ["A"] | null | $${U_n} = \prod\limits_{r = 1}^n {{{\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}^r}} $$<br><br>$$L = \mathop {\lim }\limits_{n \to \infty } {({U_n})^{ - 4/{n^2}}}$$<br><br>$$\log L = \mathop {\lim }\limits_{n \to \infty } {{ - 4} \over {{n^2}}}\sum\limits_{r = 1}^n {\log {{\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}^r}} $$<br><br>$$ \Rightarrow \log L = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n { - {{4r} \over n}.{1 \over n}\log \left( {1 + {{{r^2}} \over {{n^2}}}} \right)} $$<br><br>$$ \Rightarrow \log L = - 4\int\limits_0^1 {x\log (1 + {x^2})\,dx} $$<br><br>put 1 + x<sup>2</sup> = t<br><br>Now, 2xdx = dt<br><br>$$ = - 2\int\limits_1^2 {\log (t)dt = - 2[t\log t - t]_1^2} $$<br><br>$$ \Rightarrow \log L = - 2(2\log 2 - 1)$$<br><br>$$\therefore$$ $$L = {e^{ - 2(2\log 2 - 1)}}$$<br><br>$$ = {e^{ - 2\left( {\log \left( {{4 \over e}} \right)} \right)}}$$<br><br>$$ = {e^{\log {{\left( {{4 \over e}} \right)}^2}}}$$<br><br>$$ = {\left( {{e \over 4}} \right)^2} = {{{e^2}} \over {16}}$$ | mcq | jee-main-2021-online-27th-august-morning-shift |
1l5b84rpc | maths | definite-integration | definite-integral-as-a-limit-of-sum | <p>$$\mathop {\lim }\limits_{n \to \infty } \left( {{{{n^2}} \over {({n^2} + 1)(n + 1)}} + {{{n^2}} \over {({n^2} + 4)(n + 2)}} + {{{n^2}} \over {({n^2} + 9)(n + 3)}} + \,\,....\,\, + \,\,{{{n^2}} \over {({n^2} + {n^2})(n + n)}}} \right)$$ is equal to :</p> | [{"identifier": "A", "content": "$${\\pi \\over 8} + {1 \\over 4}{\\log _e}2$$"}, {"identifier": "B", "content": "$${\\pi \\over 4} + {1 \\over 8}{\\log _e}2$$"}, {"identifier": "C", "content": "$${\\pi \\over 4} - {1 \\over 8}{\\log _e}2$$"}, {"identifier": "D", "content": "$${\\pi \\over 8} + {\\log _e}\\sqrt 2 $$"}] | ["A"] | null | <p>$$\mathop {\lim }\limits_{n \to \infty } \left( {{{{n^2}} \over {({n^2} + 1)(n + 1)}} + {{{n^2}} \over {({n^2} + 4)(n + 2)}} + \,\,...\,\, + \,\,{{{n^2}} \over {({n^2} + {n^2})(n + n)}}} \right)$$</p>
<p>$$ = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{{{n^2}} \over {({n^2} + {r^2})(n + r)}}} $$</p>
<p>$$ = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{1 \over n}{1 \over {\left[ {1 + {{\left( {{r \over n}} \right)}^2}} \right]\left[ {1 + \left( {{r \over n}} \right)} \right]}}} $$</p>
<p>$$ = \int\limits_0^1 {{1 \over {(1 + {x^2})(1 + x)}}dx} $$</p>
<p>$$ = {1 \over 2}\int_0^1 {\left[ {{1 \over {1 + x}} - {{(x - 1)} \over {(1 + {x^2})}}} \right]dx} $$</p>
<p>$$ = {1 \over 2}\left[ {\ln (1 + x) - {1 \over 2}\ln \left( {1 + {x^2}} \right) + {{\tan }^{ - 1}}x} \right]_0^1$$</p>
<p>$$ = {1 \over 2}\left[ {{\pi \over 4} + {1 \over 2}\ln 2} \right] = {\pi \over 8} + {1 \over 4}\ln 2$$</p> | mcq | jee-main-2022-online-24th-june-evening-shift |
1l5vzsdov | maths | definite-integration | definite-integral-as-a-limit-of-sum | <p>$$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{r \over {2{r^2} - 7rn + 6{n^2}}}} $$ is equal to :</p> | [{"identifier": "A", "content": "$${\\log _e}\\left( {{{\\sqrt 3 } \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$${\\log _e}\\left( {{{3\\sqrt 3 } \\over 4}} \\right)$$"}, {"identifier": "C", "content": "$${\\log _e}\\left( {{{27} \\over 4}} \\right)$$"}, {"identifier": "D", "content": "$${\\log _e}\\left( {{4 \\over 3}} \\right)$$"}] | ["B"] | null | <p>$$\mathop {\lim }\limits_{n \to \alpha } \sum\limits_{r = 1}^n {{r \over {2{r^2} - 7rn + 6{n^2}}}} $$</p>
<p>$$ = \mathop {\lim }\limits_{n \to \alpha } {1 \over n}\sum\limits_{r = 1}^n {{{\left( {{r \over n}} \right)} \over {2{{\left( {{r \over n}} \right)}^2} - 7\left( {{r \over n}} \right) + 6}}} $$</p>
<p>$$ = \int_a^b {f(x)dx} $$</p>
<p>$$a = \mathop {\lim }\limits_{n \to \alpha } \left( {{1 \over n}} \right) = 0$$</p>
<p>$$b = \mathop {\lim }\limits_{n \to \alpha } \left( {{n \over n}} \right) = 1$$</p>
<p>and $${r \over n} \to x$$</p>
<p>$$ = \int_0^1 {{x \over {2{x^2} - 7x + 6}}dx} $$</p>
<p>$$ = \int_0^1 {{x \over {2{x^2} - 3x - 4x + 6}}dx} $$</p>
<p>$$ = \int_0^1 {{x \over {(2x - 3)(x - 2)}}dx} $$</p>
<p>$$ = \int_0^1 {\left[ {{A \over {(2x - 3)}} + {B \over {(x - 2)}}} \right]dx} $$</p>
<p>$$ = \int_0^1 {\left( {{{ - 3} \over {2x - 3}} + {2 \over {x - 2}}} \right)dx} $$</p>
<p>$$ = \left[ { - {{3\log |2x - 3|} \over 2} + 2\log |x - 2|} \right]_0^1$$</p>
<p>$$ = - {3 \over 2}\left[ {\log ( - 1) - \log ( - 3)} \right] + 2\left[ {\log ( - 1) - \log ( - 2)} \right]$$</p>
<p>$$ = - {3 \over 2}\log \left( {{1 \over 3}} \right) + 2\log \left( {{1 \over 2}} \right)$$</p>
<p>$$ = + {3 \over 2}\log 3 - 2\log 2$$</p>
<p>$$ = \log \sqrt {{3^3}} - \log 4$$</p>
<p>$$ = \log {{\sqrt {{3^2} \times 3} } \over 4}$$</p>
<p>$$ = \log {{3\sqrt 3 } \over 4}$$</p> | mcq | jee-main-2022-online-30th-june-morning-shift |
1l6dxdfr3 | maths | definite-integration | definite-integral-as-a-limit-of-sum | <p>$$
\begin{aligned}
&\text { If } \lim _{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}}[(n k+1)+(n k+2)+\ldots+(n k+n)] \\
&=33 \cdot \lim _{n \rightarrow \infty} \frac{1}{n^{k+1}} \cdot\left[1^{k}+2^{k}+3^{k}+\ldots+n^{k}\right]
\end{aligned}$$,
then the integral value of $$\mathrm{k}$$ is equal to _____________</p> | [] | null | 5 | $\lim\limits_{n \rightarrow \infty}\left(\frac{n+1}{n}\right)^{k-1} \frac{1}{n} \sum_{r=1}^{n}\left(k+\frac{r}{n}\right)=33 \lim\limits_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n}\left(\frac{r}{n}\right)^{k}$
<br/><br/>
$$
\begin{aligned}
&\Rightarrow \int_{0}^{1}(k+x) d x=33 \int_{0}^{1} x^{k} d x \\\\
&\Rightarrow \quad \frac{2 k+1}{2}=\frac{33}{k+1} \\\\
&\Rightarrow \quad k=5
\end{aligned}
$$ | integer | jee-main-2022-online-25th-july-morning-shift |
1l6f0ysr5 | maths | definite-integration | definite-integral-as-a-limit-of-sum | <p>$$\mathop {\lim }\limits_{n \to \infty } {1 \over {{2^n}}}\left( {{1 \over {\sqrt {1 - {1 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {2 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {3 \over {{2^n}}}} }} + \,\,...\,\, + \,\,{1 \over {\sqrt {1 - {{{2^n} - 1} \over {{2^n}}}} }}} \right)$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{1}{2}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$-$$2"}] | ["C"] | null | <p>$$I = \mathop {\lim }\limits_{n \to \infty } {1 \over {{2^n}}}\left( {{1 \over {\sqrt {1 - {1 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {2 \over {{2^n}}}} }} + {1 \over {\sqrt {1 - {3 \over {{2^n}}}} }} + \,\,.....\,\, + \,{1 \over {\sqrt {1 - {{{2^n} - 1} \over {{2^n}}}} }}} \right)$$</p>
<p>Let $${2^n} = t$$ and if $$n \to \infty $$ then $$t \to \infty $$</p>
<p>$$I = \mathop {\lim }\limits_{n \to \infty } {1 \over t}\left( {\sum\limits_{r = 1}^{t - 1} {{1 \over {\sqrt {1 - {r \over t}} }}} } \right)$$</p>
<p>$$I = \int\limits_0^1 {{{dx} \over {\sqrt {1 - x} }} = \int\limits_0^1 {{{dx} \over {\sqrt x }}\,\,\,\,\,\,\,\,\,\left( {\int\limits_0^a {f(x)dx = \int\limits_0^a {f(a - x)dx} } } \right.} } $$</p>
<p>$$ = \left[ {2{x^{{1 \over 2}}}} \right]_0^1 = 2$$</p> | mcq | jee-main-2022-online-25th-july-evening-shift |
1l6ghx5h6 | maths | definite-integration | definite-integral-as-a-limit-of-sum | <p>If $$a = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{{2n} \over {{n^2} + {k^2}}}} $$ and $$f(x) = \sqrt {{{1 - \cos x} \over {1 + \cos x}}} $$, $$x \in (0,1)$$, then :</p> | [{"identifier": "A", "content": "$$2\\sqrt 2 f\\left( {{a \\over 2}} \\right) = f'\\left( {{a \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$$f\\left( {{a \\over 2}} \\right)f'\\left( {{a \\over 2}} \\right) = \\sqrt 2 $$"}, {"identifier": "C", "content": "$$\\sqrt 2 f\\left( {{a \\over 2}} \\right) = f'\\left( {{a \\over 2}} \\right)$$"}, {"identifier": "D", "content": "$$f\\left( {{a \\over 2}} \\right) = \\sqrt 2 f'\\left( {{a \\over 2}} \\right)$$"}] | ["C"] | null | <p>$$a = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {{{2n} \over {{n^2} + {k^2}}}} $$</p>
<p>$$ = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{k = 1}^n {{2 \over {1 + {{\left( {{k \over n}} \right)}^2}}}} $$</p>
<p>$$a = \int\limits_0^1 {{2 \over {1 + {x^2}}}dx = 2{{\tan }^{ - 1}}x\int_0^1 { = {\pi \over 2}} } $$</p>
<p>$$f(x) = \sqrt {{{1 - \cos x} \over {1 + \cos x}}} ,\,x \in (0,1)$$</p>
<p>$$f(x) = {{1 - \cos x} \over {\sin x}} = \cos ec\,x - \cot x$$</p>
<p>$$f'(x) = \cos e{c^2}x - \cos ec\,x\cot x$$</p>
<p>$$\left. {\matrix{
{f\left( {{a \over 2}} \right) = f\left( {{\pi \over 4}} \right) = \sqrt 2 - 1} \cr
{f'\left( {{a \over 2}} \right) = f'\left( {{\pi \over 4}} \right) = 2 - \sqrt 2 } \cr
} } \right\}f'\left( {{a \over 2}} \right) = \sqrt 2 \,.\,f\left( {{a \over 2}} \right)$$</p> | mcq | jee-main-2022-online-26th-july-morning-shift |
1ldom3jmb | maths | definite-integration | definite-integral-as-a-limit-of-sum | <p>$$\mathop {\lim }\limits_{n \to \infty } \left[ {{1 \over {1 + n}} + {1 \over {2 + n}} + {1 \over {3 + n}}\, + \,...\, + \,{1 \over {2n}}} \right]$$ is equal to</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$${\\log _e}2$$"}, {"identifier": "C", "content": "$${\\log _e}\\left( {{2 \\over 3}} \\right)$$"}, {"identifier": "D", "content": "$${\\log _e}\\left( {{3 \\over 2}} \\right)$$"}] | ["B"] | null | $$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left[\frac{1}{1+n}+\frac{1}{2+n}+\frac{1}{3+n}+\ldots \ldots+\frac{1}{2 n}\right] \\\\
& =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{r+n} \\\\
& =\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{n}\left(\frac{1}{\frac{r}{n}+1}\right) \\\\
& =\int_0^1 \frac{d x}{x+1} \\\\
& =\left.\log _e(1+\mathrm{x})\right|_0 ^1 \\\\
& =\log _e^2
\end{aligned}
$$ | mcq | jee-main-2023-online-1st-february-morning-shift |
ldqvscjt | maths | definite-integration | definite-integral-as-a-limit-of-sum | $\lim\limits_{n \rightarrow \infty} \frac{3}{n}\left\{4+\left(2+\frac{1}{n}\right)^2+\left(2+\frac{2}{n}\right)^2+\ldots+\left(3-\frac{1}{n}\right)^2\right\}$ is equal to : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$\\frac{19}{3}$"}, {"identifier": "C", "content": "19"}, {"identifier": "D", "content": "12"}] | ["C"] | null | <p>$$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {{3 \over n}{{\left( {2 + {r \over n}} \right)}^2}} $$</p>
<p>$$ = \int_0^1 {3{{(2 + x)}^2}\,dx} $$</p>
<p>$$ = \left. {3\,.\,{{{{(2 + x)}^3}} \over 3}} \right|_0^1$$</p>
<p>$$ = {3^3} - {2^3} = 19$$</p> | mcq | jee-main-2023-online-30th-january-evening-shift |
1lgpxy9ft | maths | definite-integration | definite-integral-as-a-limit-of-sum | <p>Among</p>
<p>(S1): $$\lim_\limits{n \rightarrow \infty} \frac{1}{n^{2}}(2+4+6+\ldots \ldots+2 n)=1$$</p>
<p>(S2) : $$\lim_\limits{n \rightarrow \infty} \frac{1}{n^{16}}\left(1^{15}+2^{15}+3^{15}+\ldots \ldots+n^{15}\right)=\frac{1}{16}$$</p> | [{"identifier": "A", "content": "Only (S1) is true"}, {"identifier": "B", "content": "Both (S1) and (S2) are true"}, {"identifier": "C", "content": "Both (S1) and (S2) are false"}, {"identifier": "D", "content": "Only (S2) is true"}] | ["B"] | null | $$
\begin{aligned}
& S_1: \lim _{n \rightarrow \infty} \frac{1}{n^2}[2+4+6+\ldots+2 n] \\\\
& \lim _{n \rightarrow \infty} 2 \frac{n(n+1)}{2 n^2}=1 \\\\
& S_2: \lim _{n \rightarrow \infty} \frac{1}{n^{16}}\left(\sum r^{15}\right)=\lim _{n \rightarrow \infty} \frac{1}{n} \sum\left(\frac{r}{n}\right)^{15} \\\\
& =\int_0^1 x^{15} d x=\frac{1}{16} \\\\
& \therefore \text { Both } S_1 \text { and } S_2 \text { are correct. }
\end{aligned}
$$ | mcq | jee-main-2023-online-13th-april-morning-shift |
1lh2zlowl | maths | definite-integration | definite-integral-as-a-limit-of-sum | <p>Let $$f(x)=\frac{x}{\left(1+x^{n}\right)^{\frac{1}{n}}}, x \in \mathbb{R}-\{-1\}, n \in \mathbb{N}, n > 2$$.</p>
<p>If $$f^{n}(x)=\left(f \circ f \circ f \ldots .\right.$$. upto $$n$$ times) $$(x)$$, then
<br/><br/>$$\lim _\limits{n \rightarrow \infty} \int_\limits{0}^{1} x^{n-2}\left(f^{n}(x)\right) d x$$ is equal to ____________.</p> | [] | null | 0 | $$
\begin{aligned}
& \text { We have, } f(x)=\frac{x}{\left(1+x^n\right)^{1 / n}} \\\\
& \therefore f(f(x))=\frac{f(x)}{\left(1+\left[f(x)^n\right]^{1 / n}\right.}\\\\
&=\frac{\frac{x}{\left(1+x^n\right)^{1 / n}}}{\left(1+\frac{x^n}{1+x^n}\right)^{1 / n}}\\\\
&=\frac{x}{\left(1+2 x^n\right)^{1 / n}} \\\\
& f(f(f(x)))=\frac{x}{\left(1+3 x^n\right)^{1 / n}} \\\\
& \text { Similarly, } f^n(x)=\frac{x}{\left(1+n x^n\right)^{1 / n}}
\end{aligned}
$$
<br/><br/>Now, $\lim\limits_{n \rightarrow \infty} \int_0^1 x^{n-2}\left(f^n(x)\right) d x$
<br/><br/>$$
\begin{aligned}
& =\lim\limits_{n \rightarrow \infty} \int\limits_0^1 x^{n-2} \frac{x}{\left(1+n x^n\right)^{1 / n}} d x \\\\
& =\lim\limits_{n \rightarrow \infty} \int\limits_0^1 \frac{x^{n-1}}{\left(1+n x^n\right)^{1 / n}} d x
\end{aligned}
$$
<br/><br/>Let $1+n x^n=t$
<br/><br/>$$\Rightarrow n^2 x^{n-1} d x=d t$$
<br/><br/>When, $x=0$, then $t=1$
<br/><br/>When, $x=1$, then $t=1+n$
<br/><br/>$\begin{aligned} & =\lim _{n \rightarrow \infty} \int_1^{1+n} \frac{d t}{n^2(t)^{1 / n}}
=\lim _{n \rightarrow \infty} \frac{1}{n^2} \int_1^{1+n} \frac{d t}{(t)^{1 / n}} \\\\ & =\lim _{n \rightarrow \infty} \frac{1}{n^2}\left(\frac{t^{1-\frac{1}{n}}}{1-\frac{1}{n}}\right)_1^{1+n} \\\\ & =\lim _{n \rightarrow \infty} \frac{1}{n(n-1)}\left[(1+n)^{1-\frac{1}{n}}-1\right]\end{aligned}$
<br/><br/>Put $n=\frac{1}{h}$
<br/><br/>When, $n \rightarrow \infty$, then $h \rightarrow 0$
<br/><br/>$$
\begin{aligned}
& =\lim _{h \rightarrow \infty} \frac{1}{\frac{1}{h}\left(\frac{1}{h}-1\right)}\left[\left(1+\frac{1}{h}\right)^{1-h}-1\right] \\\\
& =\lim _{h \rightarrow \infty} \frac{1}{\frac{1}{h}\left(\frac{1}{h}-1\right)}\left[1-(1-h)\left(1+\frac{1}{h}\right)+\ldots-1\right] \\\\
& =\lim _{h \rightarrow 0} \frac{h^2}{1-h}\left[-(1-h)\left(1+\frac{1}{h}\right)+\ldots\right] \\\\
& =\lim _{h \rightarrow 0}-h[h+1+\ldots . .]=0
\end{aligned}
$$ | integer | jee-main-2023-online-6th-april-evening-shift |
1lsg8xaqg | maths | definite-integration | definite-integral-as-a-limit-of-sum | <p>The value of $$\lim _\limits{n \rightarrow \infty} \sum_\limits{k=1}^n \frac{n^3}{\left(n^2+k^2\right)\left(n^2+3 k^2\right)}$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{\\pi}{8(2 \\sqrt{3}+3)}$$\n"}, {"identifier": "B", "content": "$$\\frac{(2 \\sqrt{3}+3) \\pi}{24}$$\n"}, {"identifier": "C", "content": "$$\\frac{13 \\pi}{8(4 \\sqrt{3}+3)}$$\n"}, {"identifier": "D", "content": "$$\\frac{13(2 \\sqrt{3}-3) \\pi}{8}$$"}] | ["C"] | null | <p>$$\begin{aligned}
& \lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{n^3}{n^4\left(1+\frac{k^2}{n^2}\right)\left(1+\frac{3 k^2}{n^2}\right)} \\
& =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{n^3}{\left(1+\frac{k^2}{n^2}\right)\left(1+\frac{3 k^2}{n^2}\right)}
\end{aligned}$$</p>
<p>$$=\int_\limits0^1 \frac{d x}{3\left(1+x^2\right)\left(\frac{1}{3}+x^2\right)}$$</p>
<p>$$\begin{aligned}
& =\int_\limits0^1 \frac{1}{3} \times \frac{3}{2} \frac{\left(x^2+1\right)-\left(x^2+\frac{1}{3}\right)}{\left(1+x^2\right)\left(x^2+\frac{1}{3}\right)} d x \\
& =\frac{1}{2} \int_\limits0^1\left[\frac{1}{x^2+\left(\frac{1}{\sqrt{3}}\right)^2}-\frac{1}{1+x^2}\right] d x \\
& =\frac{1}{2}\left[\sqrt{3} \tan ^{-1}(\sqrt{3} x)\right]_0^1-\frac{1}{2}\left(\tan ^{-1} x\right)_0^1 \\
& =\frac{\sqrt{3}}{2}\left(\frac{\pi}{3}\right)-\frac{1}{2}\left(\frac{\pi}{4}\right)=\frac{\pi}{2 \sqrt{3}}-\frac{\pi}{8} \\
& =\frac{13 \pi}{8 \cdot(4 \sqrt{3}+3)}
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift |
luy9clal | maths | definite-integration | definite-integral-as-a-limit-of-sum | <p>Let $$\lim _\limits{n \rightarrow \infty}\left(\frac{n}{\sqrt{n^4+1}}-\frac{2 n}{\left(n^2+1\right) \sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}-\frac{8 n}{\left(n^2+4\right) \sqrt{n^4+16}}\right.$$ $$\left.+\ldots+\frac{n}{\sqrt{n^4+n^4}}-\frac{2 n \cdot n^2}{\left(n^2+n^2\right) \sqrt{n^4+n^4}}\right)$$ be $$\frac{\pi}{k}$$, using only the principal values of the inverse trigonometric functions. Then $$\mathrm{k}^2$$ is equal to _________.</p> | [] | null | 32 | <p>$$\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{n}{\sqrt{n^4+1}}+\frac{n}{\sqrt{n^4+16}}+\ldots \frac{n}{\sqrt{n^4+n^4}}\right) \\
& -\lim _{n \rightarrow \infty}\left(\frac{2 n}{\left(n^2+1\right)\left(\sqrt{n^4+1}\right)}\right)+\frac{8 n}{\left(n^2+4\right) \sqrt{n^4+1}}+\cdots \frac{2 n \cdot n^2}{\left(n^2+n^2\right) \sqrt{n^4+n^4}} \\
& =\lim _{n \rightarrow \infty} \sum_{n=1}^n \frac{1}{n \sqrt{1+\frac{r^4}{n^4}}}-\lim _{n \rightarrow \infty} \sum_{n=1}^n \frac{1}{n} \frac{2 \cdot(r / n)^2}{\left(1+\left(\frac{r}{n}\right)^2\right) \sqrt{1+\frac{r^4}{n^4}}} \\
& =\int_\limits0^1 \frac{d x}{\sqrt{1+x^4}}-2 \int_\limits0^1 \frac{x^2}{\left(1+x^2\right) \sqrt{1+x^4}} d x \\
& =\int_\limits0^1 \frac{1-x^2}{\left(1+x^2\right) \sqrt{1+x^4}} d x
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \int_\limits0^1 \frac{\left(\frac{1}{x^2}-1\right) d x}{\left(x+\frac{1}{x}\right) \sqrt{x^2+\frac{1}{x^2}}} \\
& =\int_\limits0^1 \frac{\left(\frac{1}{x^2}-1\right) d x}{\left(x+\frac{1}{x}\right) \sqrt{\left(x+\frac{1}{x}\right)^2-2}} \\
& x+\frac{1}{x}=t \Rightarrow\left(1-\frac{1}{x^2}\right) d x=d t \\
& -\int_\limits{\infty}^2 \frac{d t}{t \sqrt{t^2-2}}=\int_\limits2^{\infty} \frac{d t}{t \sqrt{t^2-2}}=\left.\frac{-1}{\sqrt{2}} \sin ^{-1} \frac{\sqrt{2}}{x}\right|_2 ^{\infty} \\
& =\frac{-1}{\sqrt{2}}\left(0-\frac{\pi}{4}\right) \\
& =\frac{\pi}{2^{5 / 2}} \\
& \therefore k=2^{5 / 2} \\
& \therefore k^2=32
\end{aligned}$$</p> | integer | jee-main-2024-online-9th-april-morning-shift |
I6qJxSWERiZM8zTy | maths | definite-integration | newton-lebnitz-rule-of-differentiation | The value of $$\mathop {\lim }\limits_{x \to 0} {{\int\limits_0^{{x^2}} {{{\sec }^2}tdt} } \over xsinx}$$ is | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}] | ["D"] | null | $$\mathop {\lim }\limits_{x \to 0} {{{d \over {dx}}\int\limits_0^{{x^2}} {{{\sec }^2}tdt} } \over {{d \over x}\left( {x\sin x} \right)}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} {{{{\sec }^2}{x^2}.2x} \over {\sin \,x + x\,\cos \,x}}$$ (by $$L'$$ Hospital rule)
<br><br>$$\mathop {\lim }\limits_{x \to 0} {{2{{\sec }^2}{x^2}} \over {\left( {{{\sin x} \over x} + \cos \,x} \right)}} = {{2 \times 1} \over {1 + 1}} = 1$$ | mcq | aieee-2003 |
pJMN8tkqVsHXcQwI | maths | definite-integration | newton-lebnitz-rule-of-differentiation | Let $$f:R \to R$$ be a differentiable function having $$f\left( 2 \right) = 6$$,
<br/>$$f'\left( 2 \right) = \left( {{1 \over {48}}} \right)$$. Then $$\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f\left( x \right)} {{{4{t^3}} \over {x - 2}}dt} $$ equals : | [{"identifier": "A", "content": "$$24$$ "}, {"identifier": "B", "content": "$$36$$ "}, {"identifier": "C", "content": "$$12$$ "}, {"identifier": "D", "content": "$$18$$ "}] | ["D"] | null | $$\mathop {\lim }\limits_{x \to 2} \int\limits_6^{f\left( x \right)} {{{4{t^3}} \over {x - 2}}} dt$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} {{\int\limits_6^{f\left( x \right)} {4{t^3}dt} } \over {x - 2}}$$
<br/><br/>This limit resembles a derivative because the fraction has the form $0/0$ as $x \to 2$ since both the numerator (integral from $6$ to $f(x)$) and the denominator ($x - 2$) are zero when $x = 2$.
<br><br>Applying $$L'$$ Hospital rule
<br><br>$$\mathop {\lim }\limits_{x \to 2} {{\left[ {4f{{\left( x \right)}^3}f'\left( x \right)} \right]} \over 1}$$
<br><br>$$ = 4{\left( {f\left( 2 \right)} \right)^3}f'\left( 2 \right)$$
<br><br>$$ = 4 \times {6^3} \times {1 \over {48}} = 18$$ | mcq | aieee-2005 |
QoEAdmndUyEMxCwIweErw | maths | definite-integration | newton-lebnitz-rule-of-differentiation | For x $$ \in $$ <b>R</b>, x $$ \ne $$ 0, if y(x) is a differentiable function such that
<br/><br/>x $$\int\limits_1^x y $$ (t) dt = (x + 1) $$\int\limits_1^x ty $$ (t) dt, then y (x) equals :
<br/><br/>(where C is a constant.) | [{"identifier": "A", "content": "$${C \\over x}{e^{ - {1 \\over x}}}$$"}, {"identifier": "B", "content": "$${C \\over {{x^2}}}{e^{ - {1 \\over x}}}$$"}, {"identifier": "C", "content": "$${C \\over {{x^3}}}{e^{ - {1 \\over x}}}$$"}, {"identifier": "D", "content": "$$C{x^3}\\,{1 \\over {{e^x}}}$$ "}] | ["C"] | null | $$x\int\limits_1^x {y\left( t \right)dt} = x\int\limits_1^x {ty} \left( t \right)dt + \int\limits_1^x {ty\left( t \right)} \,\,dt$$
<br><br>Differentiate w.r. to x.
<br><br>$$\int\limits_1^x {y\left( t \right)dt + x\left[ {y\left( x \right) - y\left( 1 \right)} \right]} $$
<br><br>$$ = \int\limits_1^x {ty\left( t \right)dt + x\left[ {xy\left( x \right) - y\left( 1 \right)} \right] + xy\left( x \right) - y\left( 1 \right)} $$
<br><br>$$\int\limits_1^x {y\left( t \right)dt = \int\limits_1^x {ty\left( t \right)dt + {x^2}y\left( x \right) - y\left( 1 \right)} } $$
<br><br>Diff. again w.r.t to x
<br><br>$$y\left( x \right) - y\left( a \right) = xy\left( x \right) - y\left( a \right) + 2x\,y\left( x \right) + {x^2}{y^3}\left( x \right)$$
<br>$$\left( {1 - 3x} \right)y\left( x \right) = {x^2}{y^3}\,\left( x \right)$$
<br><br>$${{{y^3}\left( x \right)} \over {y\left( x \right)}} = {{1 - 3x} \over {{x^2}}}$$
<br><br>$${{1dy} \over {ydx}} = {{1 - 3x} \over {{x^2}}} \Rightarrow \ln \,y = - {1 \over x} - 3\ln x$$
<br><br>$$\ln \left( {y{x^3}} \right) = - {1 \over x}$$
<br><br>$$y{x^3} = - {e^{ - 1/x}}$$
<br><br>$$y = {{{e^{ - {1 \over x}}}} \over {{x^3}}}$$
<br><br>or $$y = {{c{e^{ - {1 \over x}}}} \over {{x^3}}}$$ | mcq | jee-main-2016-online-10th-april-morning-slot |
PrYwpygyiAmYysm5OI6VH | maths | definite-integration | newton-lebnitz-rule-of-differentiation | Let f be a differentiable function from
<br/><br/><b>R</b> to <b>R</b> such that $$\left| {f\left( x \right) - f\left( y \right)} \right| \le 2{\left| {x - y} \right|^{{3 \over 2}}},$$ <br/><br/>for all $$x,y \in $$ <b>R</b>.
<br/><br/>If $$f\left( 0 \right) = 1$$
<br/><br/>then $$\int\limits_0^1 {{f^2}} \left( x \right)dx$$ is equal to : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "0"}] | ["A"] | null | $$\left| {f(x) - f(y)} \right| \le 2{\left[ {x - y} \right]^{3/2}}$$
<br><br>$$\left| {{{f(x) - f(y)} \over {x - y}}} \right| \le 2{\left| {x - y} \right|^{1/2}}$$
<br><br>$$\mathop {\lim }\limits_{y \to x} \left| {{{f(x) - f(y)} \over {x - y}}} \right| \le \mathop {\lim }\limits_{y \to x} 2{\left| {x - y} \right|^{1/2}}$$
<br><br>$$ \Rightarrow \left| {f'\left( x \right)} \right| \le 0$$ $$ \Rightarrow f'\left( x \right) = 0$$
<br><br>$$ \Rightarrow f\left( x \right) = $$ constant
<br><br>as $$f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = 1$$
<br><br>$$\int\limits_0^1 {{f^2}} \left( x \right)dx = 1$$ | mcq | jee-main-2019-online-9th-january-evening-slot |
M4m67SR6ySoaeO4fudc40 | maths | definite-integration | newton-lebnitz-rule-of-differentiation | If $$\int\limits_0^x \, $$f(t) dt = x<sup>2</sup> + $$\int\limits_x^1 \, $$ t<sup>2</sup>f(t) dt then f '$$\left( {{1 \over 2}} \right)$$ is - | [{"identifier": "A", "content": "$${{18} \\over {25}}$$"}, {"identifier": "B", "content": "$${{6} \\over {25}}$$"}, {"identifier": "C", "content": "$${{24} \\over {25}}$$"}, {"identifier": "D", "content": "$${{4} \\over {5}}$$"}] | ["C"] | null | $$\int\limits_0^x \, $$f(t) dt = x<sup>2</sup> + $$\int\limits_x^1 \, $$ t<sup>2</sup>f(t) dt f '$$\left( {{1 \over 2}} \right)$$ = ?
<br><br>Differentiate w.r.t. 'x'
<br><br>f(x) = 2x + 0 $$-$$ x<sup>2</sup> f(x)
<br><br>f(x) = $${{2x} \over {1 + {x^2}}}$$ $$ \Rightarrow $$ f '(x) = $${{\left( {1 + {x^2}} \right)2 - 2x\left( {2x} \right)} \over {{{\left( {1 + {x^2}} \right)}^2}}}$$
<br><br>f '(x) = $${{2{x^2} - 4{x^2} + 2} \over {{{\left( {1 + {x^2}} \right)}^2}}}$$
<br><br>f '$$\left( {{1 \over 2}} \right) = {{2 - 2\left( {{1 \over 4}} \right)} \over {{{\left( {1 + {1 \over 4}} \right)}^2}}} = {{\left( {{3 \over 2}} \right)} \over {{{25} \over {16}}}} = {{48} \over {50}} = {{24} \over {25}}$$ | mcq | jee-main-2019-online-10th-january-evening-slot |
7eehTp46kzo6JgEura18hoxe66ijvwu5o7l | maths | definite-integration | newton-lebnitz-rule-of-differentiation | If f : R $$ \to $$ R is a differentiable function and
f(2) = 6,<br/> then $$\mathop {\lim }\limits_{x \to 2} {{\int\limits_6^{f\left( x \right)} {2tdt} } \over {\left( {x - 2} \right)}}$$ is :- | [{"identifier": "A", "content": "2f'(2)"}, {"identifier": "B", "content": "24f'(2)"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "12f'(2)"}] | ["D"] | null | $$\mathop {\lim }\limits_{x \to 2} {{\int\limits_6^{f\left( x \right)} {2tdt} } \over {\left( {x - 2} \right)}}$$
<br><br>This is $${0 \over 0}$$ form so we use L – Hospital Rule.
<br><br>= $$\mathop {\lim }\limits_{x \to 2} {{2f\left( x \right).f'\left( x \right) - 0} \over 1}$$
<br><br>= $${2f\left( 2 \right).f'\left( 2 \right)}$$
<br><br>= 12f'(2)
<br><br><b>Note :</b> Newton - Leibnitz rule of differentiation under integration
<br><br>$${d \over {dx}}\int\limits_{\alpha \left( x \right)}^{\beta \left( x \right)} {f\left( u \right)} du$$
<br><br>= $$f\left( {\beta \left( x \right)} \right)\beta '\left( x \right) - f\left( {\alpha \left( x \right)} \right)\alpha '\left( x \right)$$ | mcq | jee-main-2019-online-9th-april-evening-slot |
rLxHSZ2R8C1omj8xb43rsa0w2w9jx6g9d50 | maths | definite-integration | newton-lebnitz-rule-of-differentiation | Let f : R $$ \to $$ R be a continuously differentiable function such that f(2) = 6 and f'(2) = $${1 \over {48}}$$. If $$\int\limits_6^{f\left( x \right)} {4{t^3}} dt$$ = (x - 2)g(x), then $$\mathop {\lim }\limits_{x \to 2} g\left( x \right)$$ is equal to : | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "36"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "24"}] | ["A"] | null | Given $$\int\limits_6^{f(x)} {4{x^3}dx} = g(x).(x - 2)$$
<br><br>$$ \Rightarrow g(x) = $$ $${{\int\limits_0^{f\left( x \right)} {4{x^3}dx} } \over {x - 2}}$$<br><br>
$$ \therefore $$ $$\mathop {\lim }\limits_{x \to 2} g(x) = $$$$\mathop {\lim }\limits_{x \to 2} {{\int\limits_0^{f\left( x \right)} {4{x^3}dx} } \over {x - 2}}$$<br><br>At x = 2 this limit is in $${0 \over 0}$$ form.
<br><br>So we can use L'Hopital's rule. Use leibniz intgral rule to differentiate the integration.
<br><br>$$ \Rightarrow \mathop {\lim }\limits_{x \to 2} {{4{f^3}(x).f'(x)} \over 1} = 4 \times {6^3} \times {1 \over {48}} = 18$$ | mcq | jee-main-2019-online-12th-april-morning-slot |
Cj07jRGhIXsCZBtHfG7k9k2k5hjthqb | maths | definite-integration | newton-lebnitz-rule-of-differentiation | $$\mathop {\lim }\limits_{x \to 0} {{\int_0^x {t\sin \left( {10t} \right)dt} } \over x}$$ is equal to | [{"identifier": "A", "content": "$$ - {1 \\over 5}$$"}, {"identifier": "B", "content": "$$ - {1 \\over 10}$$"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$$ {1 \\over 10}$$"}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to 0} {{\int_0^x {t\sin \left( {10t} \right)dt} } \over x}$$
<br><br>This is in $${0 \over 0}$$ form.
<br><br>So apply newton leibniz rule
<br><br>$$\mathop {\lim }\limits_{x \to 0} {{x.\sin \left( {10x} \right) - 0} \over 1}$$ = 0 | mcq | jee-main-2020-online-8th-january-evening-slot |
ccqUxwJrneLUb7B9yj7k9k2k5kh9wos | maths | definite-integration | newton-lebnitz-rule-of-differentiation | Let a function ƒ : [0, 5] $$ \to $$ R be continuous,
ƒ(1) = 3 and F be defined as :<br/><br/>
$$F(x) = \int\limits_1^x {{t^2}g(t)dt} $$ , where $$g(t) = \int\limits_1^t {f(u)du} $$ <br/><br/>Then for the function F, the point x = 1 is : | [{"identifier": "A", "content": "a point of inflection."}, {"identifier": "B", "content": "a point of local maxima."}, {"identifier": "C", "content": "a point of local minima."}, {"identifier": "D", "content": "not a critical point."}] | ["C"] | null | $$F(x) = \int\limits_1^x {{t^2}g(t)dt} $$
<br><br>$$ \Rightarrow $$ F'(x) = x<sup>2</sup>g(x) = x<sup>2</sup>$$\int\limits_1^t {f(u)du} $$
<br><br>$$ \therefore $$ F'(1) = (1)(0) = 0
<br><br>Now, F''(x) = 2xg(x) + x<sup>2</sup>g'(x)
<br><br>F''(1) = 2g(1) + g'(1) = 0 + g'(1) = 3
<br><br>[ As g'(t) = f(t); g'(1) = f'(1) = 3 ]
<br><br>So, at x = 1, F'(1) = 0 and F"(1) = 3 > 0
<br><br>$$ \therefore $$ For the function f(x), x = 1 is a point of local minima.
| mcq | jee-main-2020-online-9th-january-evening-slot |
eWTJ6SlzQ17mZgakyQjgy2xukfw0k3gs | maths | definite-integration | newton-lebnitz-rule-of-differentiation | $$\mathop {\lim }\limits_{x \to 1} \left( {{{\int\limits_0^{{{\left( {x - 1} \right)}^2}} {t\cos \left( {{t^2}} \right)dt} } \over {\left( {x - 1} \right)\sin \left( {x - 1} \right)}}} \right)$$ | [{"identifier": "A", "content": "is equal to 0"}, {"identifier": "B", "content": "is equal to $${1 \\over 2}$$"}, {"identifier": "C", "content": "does not exist"}, {"identifier": "D", "content": "is equal to $$ - {1 \\over 2}$$"}] | ["A"] | null | $$\mathop {\lim }\limits_{x \to 1} \left( {{{\int\limits_0^{{{\left( {x - 1} \right)}^2}} {t\cos \left( {{t^2}} \right)dt} } \over {\left( {x - 1} \right)\sin \left( {x - 1} \right)}}} \right)$$$$\left( {{0 \over 0}} \right)$$
<br><br>Apply L Hospital Rule
<br><br>= $$\mathop {\lim }\limits_{x \to 1} {{2\left( {x - 1} \right).{{\left( {x - 1} \right)}^2}\cos {{\left( {x - 1} \right)}^4} - 0} \over {\left( {x - 1} \right).\cos \left( {x - 1} \right) + \sin \left( {x - 1} \right)}}$$ $$\left( {{0 \over 0}} \right)$$
<br><br>= $$\mathop {\lim }\limits_{x \to 1} {{2{{\left( {x - 1} \right)}^3}\cos {{\left( {x - 1} \right)}^4}} \over {\left( {x - 1} \right).\left[ {\cos \left( {x - 1} \right) + {{\sin \left( {x - 1} \right)} \over {\left( {x - 1} \right)}}} \right]}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to 1} {{2{{\left( {x - 1} \right)}^2}\cos {{\left( {x - 1} \right)}^4}} \over {\cos \left( {x - 1} \right) + {{\sin \left( {x - 1} \right)} \over {\left( {x - 1} \right)}}}}$$
<br><br>on taking limit
<br><br>= $${0 \over {1 + 1}}$$ = 0 | mcq | jee-main-2020-online-6th-september-morning-slot |
O2ZShb2HjKklQdRLCL1klrfaiok | maths | definite-integration | newton-lebnitz-rule-of-differentiation | $$\mathop {\lim }\limits_{x \to 0} {{\int\limits_0^{{x^2}} {\left( {\sin \sqrt t } \right)dt} } \over {{x^3}}}$$ is equal to : | [{"identifier": "A", "content": "$${1 \\over {15}}$$"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "$${2 \\over 3}$$"}, {"identifier": "D", "content": "$${3 \\over 2}$$"}] | ["C"] | null | $$\mathop {\lim }\limits_{x \to {0^ + }} {{\int\limits_0^{{x^2}} {\sin (\sqrt t )dt} } \over {{x^3}}}$$<br><br>This is in $${0 \over 0}$$ form, so use L' Hospital rule<br><br>$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{{d \over {dx}}\left( {\int\limits_0^{{x^2}} {\sin (\sqrt t )dt} } \right)} \over {{d \over {dx}}\left( {{x^3}} \right)}}$$<br><br>$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{\sin x.2x - 0} \over {3{x^2}}}$$<br><br>(applying Leibnitz rule)<br><br>$$ = {2 \over 3}\mathop {\lim }\limits_{x \to {0^ + }} {{\sin x} \over x}$$<br><br>$$ = {2 \over 3}$$ | mcq | jee-main-2021-online-24th-february-morning-slot |
U00iEUsehx8dZ22WFB1kluwqi02 | maths | definite-integration | newton-lebnitz-rule-of-differentiation | Let $$f(x) = \int\limits_0^x {{e^t}f(t)dt + {e^x}} $$ be a differentiable function for all x$$\in$$R. Then f(x) equals : | [{"identifier": "A", "content": "$${e^{({e^{x - 1}})}}$$"}, {"identifier": "B", "content": "$$2{e^{{e^x}}} - 1$$"}, {"identifier": "C", "content": "$$2{e^{{e^x} - 1}} - 1$$"}, {"identifier": "D", "content": "$${e^{{e^x}}} - 1$$"}] | ["C"] | null | $$f(x) = \int\limits_0^x {{e^t}f(t)dt + {e^x}} $$ .... (1)<br><br>Differentiating both sides w.r.t. x<br><br>$$f'(x) = {e^x}.f(x) + {e^x}$$ (Using Newton L:eibnitz Theorem)<br><br>$$ \Rightarrow {{f'(x)} \over {f(x) + 1}} = {e^x}$$<br><br>Integrating w.r.t. x<br><br>$$\int {{{f'(x)} \over {f(x) + 1}}dx = \int {{e^x}dx} } $$<br><br>$$ \Rightarrow \ln (f(x) + 1) = {e^x} + c$$<br><br>Put x = 0<br><br>ln 2 = 1 + c ($$ \because $$ f(0) = 1, from equation (1))<br><br>$$ \therefore $$ $$\ln (f(x) + 1) = {e^x} + \ln 2 - 1$$<br><br>$$ \Rightarrow f(x) + 1 = 2.\,{e^{{e^x} - 1}}$$<br><br>$$ \Rightarrow f(x) = 2{e^{{e^x} - 1}} - 1$$ | mcq | jee-main-2021-online-26th-february-evening-slot |
kS6QZFldzuh02L2VvE1kmhzjbbd | maths | definite-integration | newton-lebnitz-rule-of-differentiation | If the normal to the curve y(x) = $$\int\limits_0^x {(2{t^2} - 15t + 10)dt} $$ at a point (a, b) is parallel to the line x + 3y = $$-$$5, a > 1, then the value of | a + 6b | is equal to ___________. | [] | null | 406 | Normal to the curve at point P(a, b) is parallel to the line x + 3y = $$-$$5.
<br><br>m<sub>normal</sub> = $$ - {1 \over 3}$$
<br><br>$$ \therefore $$ m<sub>tangent</sub> = 3 = $${{dy} \over {dx}}$$
<br><br>Given y(x) = $$\int\limits_0^x {(2{t^2} - 15t + 10)dt} $$
<br><br>$$ \Rightarrow $$ y'(x) = (2x<sup>2</sup> $$-$$ 15x + 10)<br><br>at point P(a, b)<br><br>3 = (2a<sup>2</sup> $$-$$ 15a + 10)<br><br>$$ \Rightarrow $$ 2a<sup>2</sup> $$-$$ 15a + 7 = 0<br><br>$$ \Rightarrow $$ 2a<sup>2</sup> $$-$$ 14a $$-$$ a + 7 = 0<br><br>$$ \Rightarrow $$ 2a(a $$-$$ 7) $$-$$ 1 (a $$-$$ 7) = 0<br><br>a = $${1 \over 2}$$ or 7,<br><br>given a > 1 $$ \therefore $$ a = 7<br><br>As P(a, b) lies on curve<br><br>$$ \therefore $$ $$b = \int_0^a {(2{t^2} - 15t + 10)dt} $$<br><br>$$b = \int_0^7 {(2{t^2} - 15t + 10)dt} $$<br><br>$$6b = - 413$$<br><br>$$ \therefore $$ $$|a + 6b|\, = 406$$ | integer | jee-main-2021-online-16th-march-morning-shift |
Mf5aP7idD5UI3MdU1k1kmklae28 | maths | definite-integration | newton-lebnitz-rule-of-differentiation | Let f : R $$ \to $$ R be defined as f(x) = e<sup>$$-$$x</sup>sinx. If F : [0, 1] $$ \to $$ R is a differentiable function with that F(x) = $$\int_0^x {f(t)dt} $$, then the value of $$\int_0^1 {(F'(x) + f(x)){e^x}dx} $$ lies in the interval | [{"identifier": "A", "content": "$$\\left[ {{{331} \\over {360}},{{334} \\over {360}}} \\right]$$"}, {"identifier": "B", "content": "$$\\left[ {{{330} \\over {360}},{{331} \\over {360}}} \\right]$$"}, {"identifier": "C", "content": "$$\\left[ {{{335} \\over {360}},{{336} \\over {360}}} \\right]$$"}, {"identifier": "D", "content": "$$\\left[ {{{327} \\over {360}},{{329} \\over {360}}} \\right]$$"}] | ["B"] | null | F(x) = $$\int_0^x {f(t)dt} $$
<br><br>$$ \Rightarrow $$ F'(x) = f(x) by Leibnitz theorem<br><br>I = $$\int\limits_0^1 {(F'(x) + f(x)){e^x}dx = \int\limits_0^1 {2f(x){e^x}dx} } $$<br><br>$$I = \int\limits_0^1 {2\sin x\,dx} $$<br><br>$$I = 2(1 - \cos 1)$$<br><br>$$ = 2\left\{ {1 - \left( {1 - {{{1^2}} \over {2!}} + {{{1^4}} \over {4!}} - {1 \over {6!}} + ...} \right)} \right\}$$<br><br>$$ = 2\left\{ {1 - \left( {1 - {1 \over 2} + {1 \over {24}}} \right)} \right\} < 2(1 - \cos 1) < 2\left\{ {1 - \left( {1 - {1 \over 2} + {1 \over {24}} - {1 \over {720}}} \right)} \right\}$$<br><br>$${{330} \over {360}} < 2(1 - \cos 1) < {{331} \over {360}}$$<br><br>$${{330} \over {360}} < 1 < {{331} \over {360}}$$ | mcq | jee-main-2021-online-17th-march-evening-shift |
1kryf7cd2 | maths | definite-integration | newton-lebnitz-rule-of-differentiation | Let f : (a, b) $$\to$$ R be twice differentiable function such that $$f(x) = \int_a^x {g(t)dt} $$ for a differentiable function g(x). If f(x) = 0 has exactly five distinct roots in (a, b), then g(x)g'(x) = 0 has at least : | [{"identifier": "A", "content": "twelve roots in (a, b)"}, {"identifier": "B", "content": "five roots in (a, b)"}, {"identifier": "C", "content": "seven roots in (a, b)"}, {"identifier": "D", "content": "three roots in (a, b)"}] | ["C"] | null | $$f(x) = \int_a^x {g(t)dt} $$
<br><br>$$ \Rightarrow $$ f′(x) = g(x)
<br><br>$$ \Rightarrow $$ f′'(x) = g'(x)
<br><br>Given, g(x).g'(x) = 0
<br><br>$$ \Rightarrow $$ f′(x).f′'(x) = 0
<br><br>Also given f(x) has exactly 5 roots.
<br><br>So from Rolle's theorem we can say,
<br><br>f′(x) has 4 roots and f′'(x) has 3 roots.
<br><br>$$ \therefore $$ f′(x).f′'(x) = 0 has 4 + 3 = 7 roots. | mcq | jee-main-2021-online-27th-july-evening-shift |
1ks0cs08h | maths | definite-integration | newton-lebnitz-rule-of-differentiation | Let $$F:[3,5] \to R$$ be a twice differentiable function on (3, 5) such that <br/><br/>$$F(x) = {e^{ - x}}\int\limits_3^x {(3{t^2} + 2t + 4F'(t))dt} $$. If $$F'(4) = {{\alpha {e^\beta } - 224} \over {{{({e^\beta } - 4)}^2}}}$$, then $$\alpha$$ + $$\beta$$ is equal to _______________. | [] | null | 16 | $$F(3) = 0$$<br><br>$${e^x}F(x) = \int\limits_3^x {(3{t^2} + 2t + 4F'(t))dt} $$<br><br>$${e^x}F(x) + {e^x}F'(x) = 3{x^2} + 2x + 4F'(x)$$<br><br>$$({e^x} - 4){{dy} \over {dx}} + {e^x}y = (3{x^2} + 2x)$$<br><br>$${{dy} \over {dx}} + {{{e^x}} \over {({e^x} - 4)}}y = {{(3{x^2} + 2x)} \over {({e^x} - 4)}}$$<br><br>$$y{e^{\int {{{{e^x}} \over {({e^x} - 4)}}dx} }} = \int {{{(3{x^2} + 2x)} \over {({e^x} - 4)}}{e^{\int {{{{e^x}} \over {{e^x} - 4}}dx} }}dx} $$<br><br>$$y.({e^x} - 4) = \int {(3{x^2} + 2x)dx + c} $$<br><br>$$y({e^x} - 4) = {x^3} + {x^2} + c$$<br><br>Put x = 3 $$\Rightarrow$$ c = $$-$$36<br><br>$$F(x) = {{({x^3} + {x^2} - 36)} \over {({e^x} - 4)}}$$<br><br>$$F'(x) = {{(3{x^2} + 2x)({e^x} - 4) - ({x^3} + {x^2} - 36){e^x}} \over {{{({e^x} - 4)}^2}}}$$<br><br>Now, put value of x = 4 we will get $$\alpha$$ = 12 & $$\beta$$ = 4 | integer | jee-main-2021-online-27th-july-morning-shift |
1ktiotbdz | maths | definite-integration | newton-lebnitz-rule-of-differentiation | Let f be a non-negative function in [0, 1] and twice differentiable in (0, 1). If $$\int_0^x {\sqrt {1 - {{(f'(t))}^2}} dt = \int_0^x {f(t)dt} } $$, $$0 \le x \le 1$$ and f(0) = 0, then $$\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\int_0^x {f(t)dt} $$ : | [{"identifier": "A", "content": "equals 0"}, {"identifier": "B", "content": "equals 1"}, {"identifier": "C", "content": "does not exist"}, {"identifier": "D", "content": "equals $${1 \\over 2}$$"}] | ["D"] | null | $$\int_0^x {\sqrt {1 - {{(f'(t))}^2}} dt = \int_0^x {f(t)dt} } ,\,0 \le x \le 1$$<br><br>differentiating both the sides<br><br>$$\sqrt {1 - {{(f'(x))}^2}} = f(x)$$<br><br>$$ \Rightarrow 1 - {(f'(x))^2} = {f^2}(x)$$<br><br>$${{f'(x)} \over {\sqrt {1 - {f^2}(x)} }} = 1$$<br><br>$${\sin ^{ - 1}}f(x) = x + C$$<br><br>$$\because$$ $$f(0) = 0 \Rightarrow C = 0 \Rightarrow f(x) = \sin x$$<br><br>Now, $$\mathop {\lim }\limits_{x \to 0} {{\int\limits_0^x {\sin t\,dt} } \over {{x^2}}}\left( {{0 \over 0}} \right) = {1 \over 2}$$ | mcq | jee-main-2021-online-31st-august-morning-shift |
1ktnzk3or | maths | definite-integration | newton-lebnitz-rule-of-differentiation | Let f : R $$\to$$ R be a continuous function. Then $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{{\pi \over 4}\int\limits_2^{{{\sec }^2}x} {f(x)\,dx} } \over {{x^2} - {{{\pi ^2}} \over {16}}}}$$ is equal to : | [{"identifier": "A", "content": "f (2)"}, {"identifier": "B", "content": "2f (2)"}, {"identifier": "C", "content": "2f $$\\left( {\\sqrt 2 } \\right)$$"}, {"identifier": "D", "content": "4f (2)"}] | ["B"] | null | $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{{\pi \over 4}\int\limits_2^{{{\sec }^2}x} {f(x)\,dx} } \over {{x^2} - {{{\pi ^2}} \over {16}}}}$$<br><br>= $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {\pi \over 4}.{{\left[ {f({{\sec }^2}x).2\sec x.\sec x\tan x} \right]} \over {2x}}$$<br><br>= $$\mathop {\lim }\limits_{x \to {\pi \over 4}} {\pi \over 4}f({\sec ^2}x){\sec ^3}x.{{\sin x} \over x}$$<br><br>= $${\pi \over 4}f(2).{\left( {\sqrt 2 } \right)^3}.{1 \over {\sqrt 2 }} \times {4 \over \pi }$$<br><br>= 2f (2) | mcq | jee-main-2021-online-1st-september-evening-shift |
1l56qx530 | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>If m and n respectively are the number of local maximum and local minimum points of the function $$f(x) = \int\limits_0^{{x^2}} {{{{t^2} - 5t + 4} \over {2 + {e^t}}}dt} $$, then the ordered pair (m, n) is equal to</p> | [{"identifier": "A", "content": "(3, 2)"}, {"identifier": "B", "content": "(2, 3)"}, {"identifier": "C", "content": "(2, 2)"}, {"identifier": "D", "content": "(3, 4)"}] | ["B"] | null | <p>$$f(x) = \int_0^{{x^2}} {{{{t^2} - 5t + 4} \over {2 + {e^t}}}dt} $$</p>
<p>$$f'(x) = 2x\left( {{{{x^4} - 5{x^2} + 4} \over {2 + {e^{{x^2}}}}}} \right) = 0$$</p>
<p>$$x = 0$$, or $$({x^2} - 4)({x^2} - 1) = 0$$</p>
<p>$$x = 0,$$ $$x = \pm 2,\, \pm 1$$</p>
<p>Now, $$f'(x) = {{2x(x + 1)(x - 1)(x + 2)(x - 2)} \over {\left( {{e^{{x^2}}} + 2} \right)}}$$</p>
<p>f'(x) changes sign from positive to negative at x = $$-$$1, 1 So, number of local maximum points = 2</p>
<p>f'(x) changes sign from negative to positive at x = $$-$$2, 0, 2 So, number of local minimum points = 3</p>
<p>$$\therefore$$ m = 2, n = 3</p> | mcq | jee-main-2022-online-27th-june-evening-shift |
1l56qytz0 | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>Let f be a differentiable function in $$\left( {0,{\pi \over 2}} \right)$$. If $$\int\limits_{\cos x}^1 {{t^2}\,f(t)dt = {{\sin }^3}x + \cos x} $$, then $${1 \over {\sqrt 3 }}f'\left( {{1 \over {\sqrt 3 }}} \right)$$ is equal to</p> | [{"identifier": "A", "content": "$$6 - 9\\sqrt 2 $$"}, {"identifier": "B", "content": "$$6 - {9 \\over {\\sqrt 2 }}$$"}, {"identifier": "C", "content": "$${9 \\over 2} - 6\\sqrt 2 $$"}, {"identifier": "D", "content": "$${9 \\over {\\sqrt 2 }} - 6$$"}] | ["B"] | null | <p>$$\int\limits_{\cos x}^1 {{t^2}f(t)dt = {{\sin }^3}x + \cos x} $$</p>
<p>$$ \Rightarrow \sin x{\cos ^2}x\,f(\cos x) = 3{\sin ^2}x\cos x - \sin x$$</p>
<p>$$ \Rightarrow f(\cos x) = 3\tan x - {\sec ^2}x$$</p>
<p>$$ \Rightarrow f'(\cos x).\,( - \sin x) = 3{\sec ^2}x - 2{\sec ^2}x\tan x$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5p7dl91/1ede20b3-057f-48f6-b08a-8900a3a65cc1/24ed3050-05bf-11ed-8617-d71e6444d1a0/file-1l5p7dl92.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5p7dl91/1ede20b3-057f-48f6-b08a-8900a3a65cc1/24ed3050-05bf-11ed-8617-d71e6444d1a0/file-1l5p7dl92.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th June Evening Shift Mathematics - Definite Integration Question 121 English Explanation"></p>
<p>Put $$\cos x = {1 \over {\sqrt 3 }}$$,</p>
<p>$$\therefore$$ $$f'\left( {{1 \over {\sqrt 3 }}} \right)\left( { - {{\sqrt 2 } \over {\sqrt 3 }}} \right) = 9 - 6\sqrt 2 $$</p>
<p>$${1 \over {\sqrt 3 }}f'\left( {{1 \over {\sqrt 3 }}} \right) = 6 - {9 \over {\sqrt 2 }}$$</p> | mcq | jee-main-2022-online-27th-june-evening-shift |
1l6f3imiz | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>Let $$f$$ be a twice differentiable function on $$\mathbb{R}$$. If $$f^{\prime}(0)=4$$ and $$f(x) + \int\limits_0^x {(x - t)f'(t)dt = \left( {{e^{2x}} + {e^{ - 2x}}} \right)\cos 2x + {2 \over a}x} $$, then $$(2 a+1)^{5}\, a^{2}$$ is equal to _______________.</p> | [] | null | 8 | $$
\begin{aligned}
\because f(x)+\int_0^x(x-t) f^{\prime}(t) d t & =\left(e^{2 x}+e^{-2 x}\right) \cos 2 x+\frac{2 x}{a} ~~...(i)
\end{aligned}
$$<br/><br/>
Here $f(0)=2 \hspace{0.5cm} ...(ii)$<br/><br/>
On differentiating equation (i) w.r.t. $x$ we get :<br/><br/>
$$
\begin{aligned}
& f^{\prime}(x)+\int_0^x f^{\prime}(t) d t+x f^{\prime}(x)-x f^{\prime}(x) \\\\
& = 2\left(e^{2 x}-e^{-2 x}\right) \cos 2 x-2\left(e^{2 x}+e^{-2 x}\right) \sin 2 x+\frac{2}{a} \\\\
& \Rightarrow \quad f(x)+f(x)-f(0)\\\\
&=2\left(e^{2 x}-e^{-2 x}\right) \cos 2 x-2\left(e^{2 x}+e^{-2 x}\right) \sin 2 x+\frac{2}{a} \\\\
& \quad \text { Replace } x \text { by } 0 \text { we get }: \\\\
& \Rightarrow \quad 4=\frac{2}{a} \Rightarrow a=\frac{1}{2} \cdot \\\\
& \therefore \quad(2 a+1)^5 \cdot a^2=2^5 \cdot \frac{1}{2^2}=2^3=8
\end{aligned}
$$ | integer | jee-main-2022-online-25th-july-evening-shift |
1l6klsu23 | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>Let f be a differentiable function satisfying $$f(x)=\frac{2}{\sqrt{3}} \int\limits_{0}^{\sqrt{3}} f\left(\frac{\lambda^{2} x}{3}\right) \mathrm{d} \lambda, x>0$$ and $$f(1)=\sqrt{3}$$. If $$y=f(x)$$ passes through the point $$(\alpha, 6)$$, then $$\alpha$$ is equal to _____________.</p> | [] | null | 12 | <p>$$\because$$ $$f(x) = {2 \over {\sqrt 3 }}\int\limits_0^{\sqrt 3 } {f\left( {{{{\lambda ^2}x} \over 3}} \right)d\lambda ,\,x > 0} $$</p>
<p>On differentiating both sides w.r.t., x, we get</p>
<p>$$f'(x) = {2 \over {\sqrt 3 }}\int\limits_0^{\sqrt 3 } {{{{\lambda ^2}} \over 3}f'\left( {{{{\lambda ^2}x} \over 3}} \right)d\lambda } $$</p>
<p>$$f'(x) = {1 \over {\sqrt 3 }}\int\limits_0^{\sqrt 3 } {\lambda \,.\,{{2\lambda } \over 3}f'\left( {{{{\lambda ^2}x} \over 3}} \right)d\lambda } $$</p>
<p>$$\therefore$$ $$\sqrt 3 f'(x) = \left[ {{\lambda \over x}\,.\,f\left( {{{{\lambda ^2}x} \over 3}} \right)} \right]_0^{\sqrt 3 } - \int\limits_0^{\sqrt 3 } {{1 \over x}f\left( {{{{\lambda ^2}x} \over 3}} \right)dx} $$</p>
<p>$$\sqrt 3 x\,f'(x) = \sqrt 3 f(x) - {{\sqrt 3 } \over 2}f(x)$$</p>
<p>$$xf'(x) = {{f(x)} \over 2}$$</p>
<p>On integrating we get : $$\ln y = {1 \over 2}\ln x + \ln c$$</p>
<p>$$\because$$ $$f(1) = \sqrt 3 $$ then $$c = \sqrt 3 $$</p>
<p>$$\therefore$$ ($$\alpha$$, 6) lies on</p>
<p>$$\therefore$$ $$y = \sqrt {3x} $$</p>
<p>$$\therefore$$ $$6 = \sqrt {3\alpha } \Rightarrow \alpha = 12$$.</p> | integer | jee-main-2022-online-27th-july-evening-shift |
1l6m6dywt | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>The minimum value of the twice differentiable function $$f(x)=\int\limits_{0}^{x} \mathrm{e}^{x-\mathrm{t}} f^{\prime}(\mathrm{t}) \mathrm{dt}-\left(x^{2}-x+1\right) \mathrm{e}^{x}$$, $$x \in \mathbf{R}$$, is :</p> | [{"identifier": "A", "content": "$$-\\frac{2}{\\sqrt{\\mathrm{e}}}$$"}, {"identifier": "B", "content": "$$-2 \\sqrt{\\mathrm{e}}$$"}, {"identifier": "C", "content": "$$-\\sqrt{\\mathrm{e}}$$"}, {"identifier": "D", "content": "$$\\frac{2}{\\sqrt{\\mathrm{e}}}$$"}] | ["A"] | null | <p>$$f(x) = \int\limits_0^x {{e^{x - t}}f'(t)dt - ({x^2} - x + 1){e^x}} $$</p>
<p>$$f(x) = {e^x}\int\limits_0^x {{e^{ - t}}f'(t)dt - ({x^2} - x + 1){e^x}} $$</p>
<p>$${e^{ - x}}f(x) = \int\limits_0^x {{e^{ - t}}f'(t)dt - ({x^2} - x + 1)} $$</p>
<p>Differentiate on both side</p>
<p>$${e^{ - x}}f'(x) + ( - f(x){e^{ - x}}) = {e^{ - x}}f'(x) - 2x + 1$$</p>
<p>$$f(x) = {e^x}(2x - 1)$$</p>
<p>$$f'(x) = {e^x}(2) + {e^x}(2x - 1)$$</p>
<p>$$ = {e^x}(2x + 1)$$</p>
<p>$$x = - {1 \over 2}$$</p>
<p>$$f''(x) = {e^x}(2) + (2x + 1){e^x}$$</p>
<p>$$ = {e^x}(2x + 3)$$</p>
<p>For $$x = - {1 \over 2}\,\,f''(x) > 0$$</p>
<p>$$\Rightarrow$$ Maxima</p>
<p>$$\therefore$$ Max. $$ = {e^{ - {1 \over 2}}}( - 1 - 1)$$</p>
<p>$$\therefore$$ $$ - {2 \over {\sqrt e }}$$</p> | mcq | jee-main-2022-online-28th-july-morning-shift |
ldo9n0jy | maths | definite-integration | newton-lebnitz-rule-of-differentiation | If $\phi(x)=\frac{1}{\sqrt{x}} \int\limits_{\frac{\pi}{4}}^x\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t, x>0$,
<br/><br/>then $\emptyset^{\prime}\left(\frac{\pi}{4}\right)$ is equal to : | [{"identifier": "A", "content": "$\\frac{4}{6+\\sqrt{\\pi}}$"}, {"identifier": "B", "content": "$\\frac{4}{6-\\sqrt{\\pi}}$"}, {"identifier": "C", "content": "$\\frac{8}{\\sqrt{\\pi}}$"}, {"identifier": "D", "content": "$\\frac{8}{6+\\sqrt{\\pi}}$"}] | ["D"] | null | $\phi(x)=\frac{1}{\sqrt{x}} \int_{\pi / 4}^{x}\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t$
<br/><br/>$\Rightarrow \phi^{\prime}(x)=\frac{-1}{2 x^{3 / 2}} \int_{\pi / 4}^{x}\left(4 \sqrt{2} \sin t-3 \phi^{\prime}(t)\right) d t$
<br/><br/>$+\frac{1}{\sqrt{x}}\left(4 \sqrt{2} \sin (x)-3 \phi^{\prime}(x)\right)$
<br/><br/>At, $x=\frac{\pi}{4}$
<br/><br/>$\phi^{\prime}\left(\frac{\pi}{4}\right)=\frac{-1}{2\left(\frac{\pi}{4}\right)^{3 / 2}} \times 0+\sqrt{\frac{4}{\pi}}\left(4 \sqrt{2} \times \frac{1}{\sqrt{2}}-3 \phi^{\prime}\left(\frac{\pi}{4}\right)\right)$
<br/><br/>$\Rightarrow \phi^{\prime}\left(\frac{\pi}{4}\right)\left[1+\frac{6}{\sqrt{\pi}}\right]=\frac{2}{\sqrt{\pi}} \times 4$
<br/><br/>$\Rightarrow \phi^{\prime}\left(\frac{\pi}{4}\right)=\frac{8}{6+\sqrt{x}}$ | mcq | jee-main-2023-online-31st-january-evening-shift |
1ldr7pott | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>$$\lim_\limits{x \rightarrow 0} \frac{48}{x^{4}} \int_\limits{0}^{x} \frac{t^{3}}{t^{6}+1} \mathrm{~d} t$$ is equal to ___________.</p> | [] | null | 12 | $$
48 \lim\limits_{x \rightarrow 0} \frac{\int_0^x \frac{t^3}{t^6+1} d t}{x^4}\left(\frac{0}{0}\right)
$$
<br/><br/>Applying L' Hospitals Rule
<br/><br/>$$
\begin{aligned}
48 \lim _{x \rightarrow 0} \frac{x^3}{x^6+1} \times \frac{1}{4 x^3}\\\\
\end{aligned}
$$
<p>= $${{48} \over 4}$$$$\mathop {\lim }\limits_{x \to 0} {{1} \over {{x^6} + 1}} = 12$$</p> | integer | jee-main-2023-online-30th-january-morning-shift |
1ldwxrkys | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>Let $$f$$ be $$a$$ differentiable function defined on $$\left[ {0,{\pi \over 2}} \right]$$ such that $$f(x) > 0$$ and $$f(x) + \int_0^x {f(t)\sqrt {1 - {{({{\log }_e}f(t))}^2}} dt = e,\forall x \in \left[ {0,{\pi \over 2}} \right]}$$. Then $$\left( {6{{\log }_e}f\left( {{\pi \over 6}} \right)} \right)^2$$ is equal to __________.</p> | [] | null | 27 | $f(x)+\int_{0}^{x} f(t) \sqrt{1-\left(\log _{e} f(t)\right)^{2}} d t=e\quad...(1)$
<br/><br/>
So, $f(0)=e$
<br/><br/>
Now differentiate w.r. to $x$
<br/><br/>
$$
\begin{gathered}
f^{\prime}(x)+f(x) \sqrt{1-\left(\log _{e} f(x)^{2}\right.}=0 \\\\
\frac{f^{\prime}(x)}{f(x) \sqrt{1-\left(\log _{e} f(x)\right)^{2}}}=-1
\end{gathered}
$$
<br/><br/>
Let $\log _{e} f(x)=t$
<br/><br/>
$\therefore \int \frac{d t}{\sqrt{1-t^{2}}}=-x+c$
<br/><br/>
$\Rightarrow \sin ^{-1} t=-x+c$
<br/><br/>
Now $f(0)=e \Rightarrow t=1$ So, $c=\frac{\pi}{2}$
<br/><br/>
$\therefore t=\sin \left(\frac{\pi}{2}-x\right)=\cos x \quad\left(\because x \in\left[0, \frac{\pi}{2}\right]\right)$
<br/><br/>
$\therefore\left(6 \log _{e} f\left(\frac{\pi}{6}\right)\right)^{2}=27$ | integer | jee-main-2023-online-24th-january-evening-shift |
1lgvpopd6 | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>Let $$f$$ be a continuous function satisfying $$\int_\limits{0}^{t^{2}}\left(f(x)+x^{2}\right) d x=\frac{4}{3} t^{3}, \forall t > 0$$. Then
$$f\left(\frac{\pi^{2}}{4}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "$$-\\pi\\left(1+\\frac{\\pi^{3}}{16}\\right)$$"}, {"identifier": "B", "content": "$$\\pi\\left(1-\\frac{\\pi^{3}}{16}\\right)$$"}, {"identifier": "C", "content": "$$-\\pi^{2}\\left(1+\\frac{\\pi^{2}}{16}\\right)$$"}, {"identifier": "D", "content": "$$\\pi^{2}\\left(1-\\frac{\\pi^{2}}{16}\\right)$$"}] | ["B"] | null | Given that
<br/><br/>$$
\int\limits_0^{t^2}\left(f(x)+x^2\right) d x=\frac{4}{3} t^3, \forall t>0
$$
<br/><br/>On differentiating using Leibnitz rule, we get
<br/><br/>$$
\begin{aligned}
& \left(f\left(t^2\right)+t^4\right) \times 2 t=\frac{4}{3} \times 3 t^2 \\\\
& \Rightarrow f\left(t^2\right)+t^4=2 t \\\\
& \Rightarrow f\left(t^2\right)=2 t-t^4
\end{aligned}
$$
<br/><br/>On substituting $\frac{\pi}{2}$ for $t$, we get
<br/><br/>$$
f\left(\frac{\pi^2}{4}\right)=2\left(\frac{\pi}{2}\right)-\frac{\pi^4}{16}=\pi\left(1-\frac{\pi^3}{16}\right)
$$ | mcq | jee-main-2023-online-10th-april-evening-shift |
lsan5y2d | maths | definite-integration | newton-lebnitz-rule-of-differentiation | Let $f:(0, \infty) \rightarrow \mathbf{R}$ and $\mathrm{F}(x)=\int\limits_0^x \mathrm{t} f(\mathrm{t}) \mathrm{dt}$. If $\mathrm{F}\left(x^2\right)=x^4+x^5$, then $\sum\limits_{\mathrm{r}=1}^{12} f\left(\mathrm{r}^2\right)$ is equal to ____________. | [] | null | 219 | $F(x)=\int\limits_0^x t \cdot f(t) d t$
<br/><br/>$\begin{aligned} & F'(x)=x f(x) \\\\ & F\left(x^2\right)=x^4+x^5, \quad \text { let } x^2=t \\\\ & F(t)=t^2+t^{5 / 2} \\\\ & F^{\prime}(t)=2 t+5 / 2 t^{3 / 2} \\\\ & t \cdot f(t)=2 t+5 / 2 t^{3 / 2} \\\\ & f(t)=2+5 / 2 r^{1 / 2}\end{aligned}$
<br/><br/>$\begin{aligned} & \sum_{r=1}^{12} f\left(r^2\right)=\sum_{r=1}^{12} 2+\frac{5}{2} r \\\\ & =24+5 / 2\left[\frac{12(13)}{2}\right] \\\\ & =219\end{aligned}$ | integer | jee-main-2024-online-1st-february-evening-shift |
jaoe38c1lse5vpfh | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>Let $$S=(-1, \infty)$$ and $$f: S \rightarrow \mathbb{R}$$ be defined as</p>
<p>$$f(x)=\int_\limits{-1}^x\left(e^t-1\right)^{11}(2 t-1)^5(t-2)^7(t-3)^{12}(2 t-10)^{61} d t \text {, }$$</p>
<p>Let $$\mathrm{p}=$$ Sum of squares of the values of $$x$$, where $$f(x)$$ attains local maxima on $$S$$, and $$\mathrm{q}=$$ Sum of the values of $$\mathrm{x}$$, where $$f(x)$$ attains local minima on $$S$$. Then, the value of $$p^2+2 q$$ is _________.</p> | [] | null | 27 | <p>$$\mathrm{f}^{\prime}(\mathrm{x})=\left(\mathrm{e}^{\mathrm{x}}-1\right)^{11}(2 \mathrm{x}-1)^5(\mathrm{x}-2)^7(\mathrm{x}-3)^{12}(2 \mathrm{x}-10)^{61}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsnjdht7/4e50bf09-665e-42ec-a15b-ac37fbbf6eff/783c33b0-cc2d-11ee-b20d-39b621d226e3/file-6y3zli1lsnjdht8.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsnjdht7/4e50bf09-665e-42ec-a15b-ac37fbbf6eff/783c33b0-cc2d-11ee-b20d-39b621d226e3/file-6y3zli1lsnjdht8.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 31st January Morning Shift Mathematics - Definite Integration Question 30 English Explanation"></p>
<p>Local minima at $$\mathrm{x}=\frac{1}{2}, \mathrm{x}=5$$</p>
<p>Local maxima at $$\mathrm{x}=0, \mathrm{x}=2$$</p>
<p>$$\therefore \mathrm{p}=0+4=4, \mathrm{q}=\frac{1}{2}+5=\frac{11}{2}$$</p>
<p>Then $$p^2+2 q=16+11=27$$</p> | integer | jee-main-2024-online-31st-january-morning-shift |
jaoe38c1lsf0dr2z | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>$$\mathop {\lim }\limits_{x \to {\pi \over 2}} \left( {{1 \over {{{\left( {x - {\pi \over 2}} \right)}^2}}}\int\limits_{{x^3}}^{{{\left( {{\pi \over 2}} \right)}^3}} {\cos \left( {{t^{{1 \over 3}}}} \right)dt} } \right)$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{3 \\pi^2}{4}$$\n"}, {"identifier": "B", "content": "$$\\frac{3 \\pi^2}{8}$$\n"}, {"identifier": "C", "content": "$$\\frac{3 \\pi}{4}$$\n"}, {"identifier": "D", "content": "$$\\frac{3 \\pi}{8}$$"}] | ["B"] | null | <p>Using L'hospital rule</p>
<p>$$\begin{aligned}
& =\lim _\limits{x \rightarrow \frac{\pi^{-}}{2}} \frac{0-\cos x \times 3 x^2}{2\left(x-\frac{\pi}{2}\right)} \\
& =\lim _\limits{x \rightarrow \frac{\pi^{-}}{2}} \frac{\sin \left(x-\frac{\pi}{2}\right)}{2\left(x-\frac{\pi}{2}\right)} \times \frac{3 \pi^2}{4} \\
& =\frac{3 \pi^2}{8}
\end{aligned}$$</p> | mcq | jee-main-2024-online-29th-january-morning-shift |
jaoe38c1lsfl2jrf | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>Let the slope of the line $$45 x+5 y+3=0$$ be $$27 r_1+\frac{9 r_2}{2}$$ for some $$r_1, r_2 \in \mathbb{R}$$. Then $$\lim _\limits{x \rightarrow 3}\left(\int_3^x \frac{8 t^2}{\frac{3 r_2 x}{2}-r_2 x^2-r_1 x^3-3 x} d t\right)$$ is equal to _________.</p> | [] | null | 12 | <p>According to the question,</p>
<p>$$\begin{aligned}
& 27 r_1+\frac{9 r_2}{2}=-9 \\
& \lim _\limits{x \rightarrow 3} \frac{\int_\limits3^x 8 t^2 d t}{\frac{3 r_2 x}{2}-r_2 x^2-r_1 x^3-3 x} \\
& =\lim _\limits{x \rightarrow 3} \frac{8 x^2}{\frac{3 r_2^2}{2}-2 r_2 x-3 r_1 x^2-3} \text { (using LH' Rule) } \\
& =\frac{72}{\frac{3 r_2}{2}-6 r_2-27 r_1-3} \\
& =\frac{72}{-\frac{9 r_2}{2}-27 r_1-3} \\
& =\frac{72}{9-3}=12
\end{aligned}$$</p> | integer | jee-main-2024-online-29th-january-evening-shift |
1lsg47bl6 | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>Let $$\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$$ be defined as $$f(x)=a e^{2 x}+b e^x+c x$$. If $$f(0)=-1, f^{\prime}\left(\log _e 2\right)=21$$ and $$\int_0^{\log _e 4}(f(x)-c x) d x=\frac{39}{2}$$, then the value of $$|a+b+c|$$ equals</p> | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "10"}] | ["C"] | null | <p>$$\begin{array}{ll}
\mathrm{f}(\mathrm{x})=a \mathrm{e}^{2 \mathrm{x}}+b \mathrm{e}^{\mathrm{x}}+\mathrm{cx} & \mathrm{f}(0)=-1 \\\\
& \mathrm{a}+\mathrm{b}=-1 \\\\
\mathrm{f}^{\prime}(\mathrm{x})=2 a \mathrm{e}^{2 \mathrm{x}}+b \mathrm{e}^{\mathrm{x}}+\mathrm{c} & \mathrm{f}^{\prime}(\ln 2)=21 \\\\
& 8 \mathrm{a}+2 \mathrm{~b}+\mathrm{c}=21 \\\\
\int_\limits0^{\ln 4}\left(a \mathrm{e}^{2 \mathrm{x}}+b \mathrm{e}^{\mathrm{x}}\right) \mathrm{dx}=\frac{39}{2} &
\end{array}$$</p>
<p>$$\begin{aligned}
& {\left[\frac{\mathrm{ae}^{2 \mathrm{x}}}{2}+\mathrm{be}^{\mathrm{x}}\right]_0^{\ln 4}=\frac{39}{2} \Rightarrow 8 \mathrm{a}+4 \mathrm{~b}-\frac{\mathrm{a}}{2}-\mathrm{b}=\frac{39}{2}} \\\\
& 15 a+6 b=39 \\\\
& 15 a-6 a-6=39 \\\\
& 9 \mathrm{a}=45 \Rightarrow \mathrm{a}=5 \\\\
& b=-6 \\\\
& c=21-40+12=-7 \\\\
& \mathrm{a}+\mathrm{b}+\mathrm{c}-8 \\\\
& |\mathrm{a}+\mathrm{b}+\mathrm{c}|=8 \\\\
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-evening-shift |
luxweovm | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>$$\lim _\limits{x \rightarrow \frac{\pi}{2}}\left(\frac{\int_{x^3}^{(\pi / 2)^3}\left(\sin \left(2 t^{1 / 3}\right)+\cos \left(t^{1 / 3}\right)\right) d t}{\left(x-\frac{\pi}{2}\right)^2}\right)$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{3 \\pi^2}{2}$$\n"}, {"identifier": "B", "content": "$$\\frac{9 \\pi^2}{8}$$\n"}, {"identifier": "C", "content": "$$\\frac{5 \\pi^2}{9}$$\n"}, {"identifier": "D", "content": "$$\\frac{11 \\pi^2}{10}$$"}] | ["B"] | null | <p>$$\lim _\limits{x \rightarrow \frac{\pi}{2}}\left(\frac{\int_{x^3}^{(\pi / 2)^3}\left(\sin \left(2 t^{1 / 3}\right)+\cos \left(t^{1 / 3}\right)\right) d t}{\left(x-\frac{\pi}{2}\right)^2}\right)$$</p>
<p>Using Newton Leibniz theorem</p>
<p>$$\begin{aligned}
& =\lim _\limits{x \rightarrow \frac{\pi}{2}}\left[\frac{\sin \left(2 \times \frac{\pi}{2}\right) \cdot 0-\sin (2 x) \cdot 3 x^2+\left(\cos \frac{\pi}{2}\right) \cdot 0-\cos x \cdot 3 x^2}{2\left(x-\frac{\pi}{2}\right)}\right] \\
& =\lim _\limits w{x \rightarrow \frac{\pi}{2}} \frac{-3 x^2 \sin 2 x-3 x^2 \cos x}{2\left(x-\frac{\pi}{2}\right)}\left(\frac{0}{0}\right) \text { form } \\
& =\lim _\limits{x \rightarrow \frac{\pi}{2}}\left[\frac{-6 x \sin 2 x-6 x^2 \cos 2 x-6 x \cos x+3 x^2 \sin x}{2}\right] \\
& =\frac{6 \times \frac{\pi^2}{4}+3 \times \frac{\pi^2}{4}}{2} \\
& =\frac{9 \pi^2}{8}
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-evening-shift |
lv2er436 | maths | definite-integration | newton-lebnitz-rule-of-differentiation | <p>Let $$f(x)=\int_0^x\left(t+\sin \left(1-e^t\right)\right) d t, x \in \mathbb{R}$$. Then, $$\lim _\limits{x \rightarrow 0} \frac{f(x)}{x^3}$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{1}{6}$$\n"}, {"identifier": "B", "content": "$$-\\frac{1}{6}$$\n"}, {"identifier": "C", "content": "$$\\frac{2}{3}$$\n"}, {"identifier": "D", "content": "$$-\\frac{2}{3}$$"}] | ["B"] | null | <p>Given $$f(x)=\int_\limits0^x\left(t+\sin \left(1-e^t\right)\right) d t$$</p>
<p>Now, $$\lim _\limits{x \rightarrow 0} \frac{f(x)}{x^3}\left(\frac{0}{0} \text { form }\right)$$</p>
<p>$$\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\int_\limits0^x\left(t+\sin \left(1-e^t\right)\right) d t}{x^3} \\
& =\lim _{x \rightarrow 0} \frac{x+\sin \left(1-e^x\right)}{3 x^2}\left(\frac{0}{0}\right) \\
& =\lim _{x \rightarrow 0} \frac{1+\cos \left(1-e^x\right)\left(-e^x\right)}{6 x}\left(\frac{0}{0}\right) \\
& =\lim _{x \rightarrow 0} \frac{-\sin \left(1-e^x\right)\left(e^x\right)^2+\cos \left(1-e^x\right)\left(-e^x\right)}{6} \\
& =-\frac{1}{6}
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-evening-shift |
hqF2I8KjE6uVCfhw | maths | definite-integration | properties-of-definite-integration | $${I_n} = \int\limits_0^{\pi /4} {{{\tan }^n}x\,dx} $$ then $$\,\mathop {\lim }\limits_{n \to \infty } \,n\left[ {{I_n} + {I_{n + 2}}} \right]$$ equals | [{"identifier": "A", "content": "$${1 \\over 2}$$ "}, {"identifier": "B", "content": "$$1$$"}, {"identifier": "C", "content": "$$\\infty $$ "}, {"identifier": "D", "content": "zero"}] | ["B"] | null | $${I_n} + {I_{n + 2}}$$
<br><br>$$ = \int\limits_0^{\pi /4} {{{\tan }^n}} \,x\left( {1 + {{\tan }^2}x} \right)dx$$
<br><br>$$ = \int\limits_0^{\pi /4} {{{\tan }^n}} x\,{\sec ^2}x\,dx$$
<br><br>$$ = \left[ {{{{{\tan }^{n + 1}}x} \over {n + 1}}} \right]_0^{\pi /4}$$
<br><br>$$ = {{1 - 0} \over {n + 1}} = {1 \over {n + 1}}$$
<br><br>$$\therefore$$ $${I_n} + {I_{n + 2}} = {1 \over {n + 1}}$$
<br><br>$$ \Rightarrow \mathop {\lim }\limits_{n \to \infty } n\left[ {{I_n} + {I_{n + 2}}} \right]$$
<br><br>$$ = \mathop {\lim }\limits_{n \to \infty } \,n.{1 \over {n + 1}} = \mathop {\lim }\limits_{n \to \infty } {n \over {n + 1}}$$
<br><br>$$ = \mathop {\lim }\limits_{n \to \infty } {n \over {n\left( {1 + {1 \over n}} \right)}} = 1$$ | mcq | aieee-2002 |
fOTnK3XKAyiYtvBR | maths | definite-integration | properties-of-definite-integration | $$\int\limits_0^{10\pi } {\left| {\sin x} \right|dx} $$ is | [{"identifier": "A", "content": "$$20$$"}, {"identifier": "B", "content": "$$8$$"}, {"identifier": "C", "content": "$$10$$"}, {"identifier": "D", "content": "$$18$$"}] | ["A"] | null | $$I = \int\limits_0^{10\pi } {\left| {\sin x} \right|} dx$$
<br><br>$$ = 10\int\limits_0^\pi {\left| {\sin x\,} \right|} \,dx$$
<br><br>$$ = 10\int\limits_0^\pi {\sin \,x\,dx} $$
<br><br>$$\left[ \, \right.$$ as $$\left| {\sin x} \right|$$ is periodic with period $$\pi $$
<br><br>and $$sin$$ $$x > 0$$ if $$0 < x$$ $$ < \pi $$ $$\left. \, \right]$$
<br><br>$$I = 20\int\limits_0^{\pi /2} {\sin x} \,dx$$
<br><br>$$ = 20\left[ { - \cos \,x_0^{\pi /2}} \right] = 20$$ | mcq | aieee-2002 |
O2Ykyj1nWTzQNYTw | maths | definite-integration | properties-of-definite-integration | $$\int\limits_0^2 {\left[ {{x^2}} \right]dx} $$ is | [{"identifier": "A", "content": "$$2 - \\sqrt 2 $$ "}, {"identifier": "B", "content": "$$2 + \\sqrt 2 $$"}, {"identifier": "C", "content": "$$\\,\\sqrt 2 - 1$$ "}, {"identifier": "D", "content": "$$ - \\sqrt 2 - \\sqrt 3 + 5$$ "}] | ["D"] | null | $$\int\limits_0^2 {\left[ {{x^2}} \right]} dx = \int\limits_0^1 {\left[ {{x^2}} \right]dx} + \int\limits_1^{\sqrt 2 } {\left[ {{x^2}} \right]} dx + $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$$$\int\limits_{\sqrt 2 }^{\sqrt 3 } {\left[ {{x^2}} \right]} + \int\limits_{\sqrt 3 }^2 {\left[ {{x^2}} \right]} dx$$
<br><br>$$ = \int\limits_0^1 {0dx} + \int\limits_1^{\sqrt 2 } {1dx} + \int\limits_{\sqrt 2 }^{\sqrt 3 } {2dx} + \int\limits_{\sqrt 3 }^2 {3dx} $$
<br><br>$$ = \left[ x \right]_1^{\sqrt n } + \left[ {2x} \right]_{\sqrt 2 }^{\sqrt 3 } + \left[ {3x} \right]_{\sqrt 3 }^2$$
<br><br>$$ = \sqrt 2 - 1 + 2\sqrt 3 - 2\sqrt 2 + 6 - 3\sqrt 3 $$
<br><br>$$ = 5 - \sqrt 3 - \sqrt 2 $$ | mcq | aieee-2002 |
e87VKbcXQlihzLai | maths | definite-integration | properties-of-definite-integration | $$\int_{ - \pi }^\pi {{{2x\left( {1 + \sin x} \right)} \over {1 + {{\cos }^2}x}}} dx$$ is | [{"identifier": "A", "content": "$${{{\\pi ^2}} \\over 4}$$ "}, {"identifier": "B", "content": "$${{\\pi ^2}}$$ "}, {"identifier": "C", "content": "zero "}, {"identifier": "D", "content": "$${\\pi \\over 2}$$ "}] | ["B"] | null | $$\int_{ - \pi }^\pi {{{2x\left( {1 + \sin \,x} \right)} \over {1 + {{\cos }^2}x}}} dx$$
<br><br>$$ = \int_{ - \pi }^\pi {{{2x\,dx} \over {1 + {{\cos }^2}x}} + 2\int_{ - \pi }^\pi {{{x\,\sin x} \over {1 + {{\cos }^2}x}}} } dx$$
<br><br>$$ = 0 + 4\int_0^\pi {{{x\sin x\,dx} \over {1 + {{\cos }^2}x}}} ;$$
<br><br>$$\left[ \, \right.$$ as $$\int\limits_{ - a}^a {f\left( x \right)} dx = 0$$ $$\left. \, \right]$$
<br><br>if $$f(x)$$ is odd
<br><br>$$ = 2\int\limits_0^a {f\left( x \right)} dx$$ if $$f(x)$$ is even.
<br><br>$$I = 4\int_0^\pi {{{\left( {\pi - x} \right)\sin \left( {\pi - x} \right)} \over {1 + {{\cos }^2}\left( {\pi - x} \right)}}} dx$$
<br><br>$$I = 4\int_0^\pi {{{\left( {\pi - x} \right)\sin \,x} \over {1 + {{\cos }^2}x}}} $$
<br><br>$$ \Rightarrow I = 4\pi \int_0^\pi {{{\sin x\,dx} \over {1 + {{\cos }^2}x}}} - 4\int {{{x\sin x\,dx} \over {1 + {{\cos }^2}x}}} $$
<br><br>$$ \Rightarrow 2I = 4\pi \int_0^\pi {{{\sin x} \over {1 + {{\cos }^2}x}}} dx$$
<br><br>put $$\cos x = t \Rightarrow - \sin xdx = dt$$
<br><br>$$\therefore$$ $$I = - 2\pi \int\limits_1^{ - 1} {{1 \over {1 + {t^2}}}} dt$$
<br><br>$$ = 2\pi \int\limits_{ - 1}^1 {{1 \over {1 + {t^2}}}} dt$$
<br><br>$$ = 2\pi \left[ {{{\tan }^{ - 1}}t} \right]_{ - 1}^1$$
<br><br>$$ = 2\pi \left[ {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}\left( { - 1} \right)} \right]$$
<br><br>$$ = 2\pi \left[ {{\pi \over 4} - \left( {{{ - \pi } \over 4}} \right)} \right] = 2\pi {\pi \over 2} = {\pi ^2}$$ | mcq | aieee-2002 |
Vah0GQp8qnQsrnOi | maths | definite-integration | properties-of-definite-integration | If $$y=f(x)$$ makes +$$ve$$ intercept of $$2$$ and $$0$$ unit on $$x$$ and $$y$$ axes and encloses an area of $$3/4$$ square unit with the axes then $$\int\limits_0^2 {xf'\left( x \right)dx} $$ is | [{"identifier": "A", "content": "$$3/2$$ "}, {"identifier": "B", "content": "$$1$$ "}, {"identifier": "C", "content": "$$5/4$$ "}, {"identifier": "D", "content": "$$-3/4$$ "}] | ["D"] | null | We have $$\int\limits_0^2 {f\left( x \right)} dx = {3 \over 4};Now,$$
<br><br>$$\int\limits_0^2 {xf'\left( x \right)} dx$$
<br><br>$$ = x\int\limits_0^2 {f'\left( x \right)dx} - \int\limits_0^2 {f\left( x \right)} dx$$
<br><br>$$ = \left[ {x\,f\left( x \right)} \right]_0^2 - {3 \over 4}$$
<br><br>$$ = 2f\left( 2 \right) - {3 \over 4}$$
<br><br>$$ = 0 - {3 \over 4}$$
<br><br>$$\left( {} \right.$$ As $$f\left( 2 \right) = 0$$ $$\left. {} \right)$$ $$ = - {3 \over 4}.$$ | mcq | aieee-2002 |
tiedgbvfIFzGtJ0O | maths | definite-integration | properties-of-definite-integration | If $$f\left( y \right) = {e^y},$$ $$g\left( y \right) = y;y > 0$$ and
<br/><br>$$F\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)dy,} $$ then :</br> | [{"identifier": "A", "content": "$$F\\left( t \\right) = t{e^{ - t}}$$ "}, {"identifier": "B", "content": "$$F\\left( t \\right) = 1t - t{e^{ - 1}}\\left( {1 + t} \\right)$$ "}, {"identifier": "C", "content": "$$F\\left( t \\right) = {e^t} - \\left( {1 + t} \\right)$$ "}, {"identifier": "D", "content": "$$F\\left( t \\right) = t{e^t}$$."}] | ["C"] | null | $$F\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)} dy$$
<br><br>$$ = \int\limits_0^t {{e^{t - y}}} ydy = {e^t}\int\limits_0^t {{e^{ - y}}} \,ydy$$
<br><br>$$ = {e^t}\left[ { - y{e^{ - y}} - {e^{ - y}}} \right]_0^t$$
<br><br>$$ = - {e^t}\left[ {y{e^{ - y}} + {e^{ - y}}} \right]_0^t$$
<br><br>$$ = - {e^t}\left[ {t\,{e^{ - t}} + {e^{ - t}} - 0 - 1} \right]$$
<br><br>$$ = - {e^t}\left[ {{{t + 1 - {e^t}} \over {{e^t}}}} \right]$$
<br><br>$$ = {e^t} - \left( {1 + t} \right)$$ | mcq | aieee-2003 |
5d1PWQJ8XGbfoKZR | maths | definite-integration | properties-of-definite-integration | The value of the integral $$I = \int\limits_0^1 {x{{\left( {1 - x} \right)}^n}dx} $$ is | [{"identifier": "A", "content": "$${1 \\over {n + 1}} + {1 \\over {n + 2}}$$ "}, {"identifier": "B", "content": "$${1 \\over {n + 1}}$$ "}, {"identifier": "C", "content": "$${1 \\over {n + 2}}$$ "}, {"identifier": "D", "content": "$${1 \\over {n + 1}} - {1 \\over {n + 2}}$$"}] | ["D"] | null | $$I = \int\limits_0^1 {x{{\left( {1 - x} \right)}^n}} dx$$
<br><br>$$ = \int\limits_0^1 {\left( {1 - x} \right){{\left( {1 - 1 + x} \right)}^n}dx} $$
<br><br>$$ = \int\limits_0^1 {\left( {1 - x} \right)} {x^n}dx$$
<br><br>$$ = \left[ {{{{x^{n + 1}}} \over {n + 1}} - {{{x^{n + 2}}} \over {n + 2}}} \right]_0^1$$
<br><br>$$ = {1 \over {n + 1}} - {1 \over {n + 2}}$$ | mcq | aieee-2003 |
0Mv321urQHCrm7sB | maths | definite-integration | properties-of-definite-integration | Let $$f(x)$$ be a function satisfying $$f'(x)=f(x)$$ with $$f(0)=1$$ and $$g(x)$$ be a function that satisfies $$f\left( x \right) + g\left( x \right) = {x^2}$$. Then the value of the integral $$\int\limits_0^1 {f\left( x \right)g\left( x \right)dx,} $$ is | [{"identifier": "A", "content": "$$e + {{{e^2}} \\over 2} + {5 \\over 2}$$ "}, {"identifier": "B", "content": "$$e - {{{e^2}} \\over 2} - {5 \\over 2}$$"}, {"identifier": "C", "content": "$$e + {{{e^2}} \\over 2} - {3 \\over 2}$$ "}, {"identifier": "D", "content": "$$e - {{{e^2}} \\over 2} - {3 \\over 2}$$ "}] | ["D"] | null | Given $$f'\left( x \right) = f\left( x \right) \Rightarrow {{f'\left( x \right)} \over {f\left( x \right)}} = 1$$
<br><br>Integrating log
<br><br>$$f\left( x \right) = x + c \Rightarrow f\left( x \right) = {e^{x + c}}$$
<br><br>$$f\left( 0 \right) = 1 \Rightarrow f\left( x \right) = {e^x}$$
<br><br>$$\therefore$$ $$\int\limits_0^1 {f\left( x \right)g\left( x \right)} dx$$
<br><br>$$ = \int\limits_0^1 {{e^x}} \left( {{x^2} - {e^x}} \right)dx$$
<br><br>$$ = \int\limits_0^1 {{x^2}} {e^x}dx - \int\limits_0^1 {{e^{2x}}} dx$$
<br><br>$$ = \left[ {{x^2}{e^x}} \right]_0^1 - 2\left[ {x{e^x} - {e^x}} \right]_0^1 - {1 \over {20}}\left[ {{e^{2x}}} \right]_0^1$$
<br><br>$$ = e - \left[ {{{{e^2}} \over 2} - {1 \over 2}} \right] - 2\left[ {e - e + 1} \right]$$
<br><br>$$ = e - {{{e^2}} \over 2} - {3 \over 2}$$ | mcq | aieee-2003 |
YkVDaKdcUs4nswLz | maths | definite-integration | properties-of-definite-integration | If $$f\left( {a + b - x} \right) = f\left( x \right)$$ then $$\int\limits_a^b {xf\left( x \right)dx} $$ is equal to | [{"identifier": "A", "content": "$${{a + b} \\over 2}\\int\\limits_a^b {f\\left( {a + b + x} \\right)dx} $$ "}, {"identifier": "B", "content": "$${{a + b} \\over 2}\\int\\limits_a^b {f\\left( {b - x} \\right)dx} $$ "}, {"identifier": "C", "content": "$${{a + b} \\over 2}\\int\\limits_a^b {f\\left( x \\right)dx} $$ "}, {"identifier": "D", "content": "$$\\,{{b - a} \\over 2}\\int\\limits_a^b {f\\left( x \\right)dx} $$ "}] | ["C"] | null | $$I = \int\limits_a^b {xf\left( x \right)} dx$$
<br><br>$$ = \int\limits_a^b {\left( {a + b - x} \right)} f\left( {a + b - x} \right)dx$$
<br><br>$$ = \left( {a + b} \right)\int\limits_a^b {f\left( {a + b - x} \right)} dx - \int\limits_a^b {xf} \left( {a + b - x} \right)dx$$
<br><br>$$ = \left( {a + b} \right)\int\limits_a^b {f\left( x \right)dx} - \int\limits_a^b {xf\left( x \right)dx} $$
<br><br>$$\left[ {} \right.$$ As given that $$f\left( {a + b - x} \right) = f\left( x \right)$$ $$\left. {} \right]$$
<br><br>$$2I = \left( {a + b} \right)\int\limits_a^b {f\left( x \right)} dx$$
<br><br>$$ \Rightarrow I = {{\left( {a + b} \right)} \over 2}\int\limits_a^b {f\left( x \right)} dx$$ | mcq | aieee-2003 |
CGM65kHJRMbNg3A6 | maths | definite-integration | properties-of-definite-integration | The value of $$I = \int\limits_0^{\pi /2} {{{{{\left( {\sin x + \cos x} \right)}^2}} \over {\sqrt {1 + \sin 2x} }}dx} $$ is | [{"identifier": "A", "content": "$$3$$ "}, {"identifier": "B", "content": "$$1$$"}, {"identifier": "C", "content": "$$2$$"}, {"identifier": "D", "content": "$$0$$"}] | ["C"] | null | $$I = \int\limits_0^{{\pi \over 2}} {{{{{\left( {\sin \,x + \cos x} \right)}^2}} \over {\sqrt {1 + \sin 2x} }}} dx$$
<br><br>We know $$\left[ {{{\left( {\sin x + \cos x} \right)}^2} = 1 + \sin 2x} \right],\,$$ So
<br><br>$$I = \int\limits_0^{{\pi \over 2}} {{{{{\left( {\sin x + \cos x} \right)}^2}} \over {\left( {\sin x + \cos x} \right)}}} dx$$
<br><br>$$ = \int\limits_0^{{\pi \over 2}} {\left( {\sin x + \cos x} \right)dx} $$
<br><br>$$\left[ {} \right.$$ $$\sin x + \cos x > 0$$ $$\,\,if\,\,0 < x < {\pi \over 2}$$ $$\left. {} \right]$$
<br><br>or $$I = \left[ { - \cos x + \sin x} \right]_0^{{\pi \over 2}} $$
<br><br>= - cos $${\pi \over 2}$$ + sin $${\pi \over 2}$$ + cos 0 - sin 0
<br><br>= - 0 + 1 + 1 - 0
<br><br>= 2 | mcq | aieee-2004 |
yxxHHZHxwFw8kp9m | maths | definite-integration | properties-of-definite-integration | The value of $$\int\limits_{ - 2}^3 {\left| {1 - {x^2}} \right|dx} $$ is | [{"identifier": "A", "content": "$${1 \\over 3}$$ "}, {"identifier": "B", "content": "$${14 \\over 3}$$"}, {"identifier": "C", "content": "$${7 \\over 3}$$"}, {"identifier": "D", "content": "$${28 \\over 3}$$"}] | ["D"] | null | $$\int\limits_{ - 2}^3 {\left| {1 - {x^2}} \right|} dx = \int\limits_{ - 2}^3 {\left| {{x^2} - 1} \right|} dx$$
<br><br>Now $$\left| {{x^2} - 1} \right| = \left\{ {\matrix{
{{x^2} - 1} & {if} & {x \le - 1} \cr
{1 - {x^2}} & {if} & { - 1 \le x \le 1} \cr
{{x^2} - 1} & {if} & {x \ge 1} \cr
} } \right.$$
<br><br>$$\therefore$$ Integral is
<br><br>$$\int\limits_{ - 2}^{ - 1} {\left( {{x^2} - 1} \right)dx + \int\limits_{ - 1}^1 {\left( {1 - {x^2}} \right)} } dx + \int\limits_1^3 {\left( {{x^2} - 1} \right)dx} $$
<br><br>$$\left[ {{{{x^3}} \over 3} - x} \right]_{ - 2}^{ - 1} + \left[ {x - {{{x^3}} \over 3}} \right]_{ - 1}^1 + \left[ {{{{x^3}} \over 3} - x} \right]_1^3$$
<br><br>$$ = \left( { - {1 \over 3} + 1} \right) - \left( { - {8 \over 3} + 2} \right) + \left( {2 - {2 \over 3}} \right) + \left( {{{27} \over 3} - 3} \right) - \left( {{1 \over 3} - 1} \right)$$
<br><br>$$ = {2 \over 3} + {2 \over 3} + {4 \over 3} + 6 + {2 \over 3} = {{28} \over 3}$$ | mcq | aieee-2004 |
mF2HI0v3oqpZ6K8Y | maths | definite-integration | properties-of-definite-integration | If $$\int\limits_0^\pi {xf\left( {\sin x} \right)dx = A\int\limits_0^{\pi /2} {f\left( {\sin x} \right)dx,} } $$ then $$A$$ is | [{"identifier": "A", "content": "$$2\\pi $$ "}, {"identifier": "B", "content": "$$\\pi $$ "}, {"identifier": "C", "content": "$${\\pi \\over 4}$$ "}, {"identifier": "D", "content": "$$0$$"}] | ["B"] | null | Let $$I = \int\limits_0^\pi {xf\left( {\sin x} \right)} dx$$
<br><br>$$ = \int\limits_0^\pi {\left( {\pi - x} \right)} f\left( {\sin x} \right)dx$$
<br><br>$$\therefore$$ $$2I = \pi \int\limits_2^\pi {f\left( {\sin x} \right)} dx$$
<br><br>$$ = \pi .2\int\limits_0^{{\pi \over 2}} {f\left( {\sin x} \right)} dx$$
<br><br>$$\therefore$$ $$I = \pi \int\limits_0^{{\pi \over 2}} {f\left( {\sin x} \right)} dx$$
<br><br>$$ \Rightarrow A = \pi $$ | mcq | aieee-2004 |
tPVNHNIpqwfMq5M6 | maths | definite-integration | properties-of-definite-integration | If $$f\left( x \right) = {{{e^x}} \over {1 + {e^x}}},{I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}dx} $$
<br/>and $${I_2} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {g\left\{ {x\left( {1 - x} \right)} \right\}dx} ,$$ then the value of $${{{I_2}} \over {{I_1}}}$$ is | [{"identifier": "A", "content": "$$1$$"}, {"identifier": "B", "content": "$$-3$$ "}, {"identifier": "C", "content": "$$-1$$"}, {"identifier": "D", "content": "$$2$$"}] | ["D"] | null | $$f\left( x \right) = {{{e^x}} \over {1 + {e^x}}}$$
<br><br>$$ \Rightarrow f\left( { - x} \right) = {{{e^{ - x}}} \over {1 + {e^{ - x}}}} = {1 \over {{e^x} + 1}}$$
<br><br>$$\therefore$$ $$f\left( x \right) + f\left( { - x} \right) = 1\forall x$$
<br><br>Now $${I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}} dx$$
<br><br>$$ = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {\left( {1 - x} \right)} g\left\{ {x\left( {1 - x} \right)} \right\}dx$$
<br><br>$$\left[ {} \right.$$ using $$\int\limits_a^b {f\left( x \right)} dx\,a$$
<br><br>$$ = \int\limits_a^b {f\left( {a + b - x} \right)dx} $$ $$\left. \, \right]$$
<br><br>$$ = {I_2} - {I_1} \Rightarrow 2{I_1} = {I_2}$$
<br><br>$$\therefore$$ $${{{I_2}} \over {{I_1}}} = 2$$ | mcq | aieee-2004 |
WZH8LobJQ9h9MXPm | maths | definite-integration | properties-of-definite-integration | If $${I_1} = \int\limits_0^1 {{2^{{x^2}}}dx,{I_2} = \int\limits_0^1 {{2^{{x^3}}}dx,\,{I_3} = \int\limits_1^2 {{2^{{x^2}}}dx} } } $$ and $${I_4} = \int\limits_1^2 {{2^{{x^3}}}dx} $$ then | [{"identifier": "A", "content": "$${I_2} > {I_1}$$ "}, {"identifier": "B", "content": "$${I_1} > {I_2}$$"}, {"identifier": "C", "content": "$${I_3} = {I_4}$$"}, {"identifier": "D", "content": "$${I_3} > {I_4}$$"}] | ["B"] | null | $${I_1} = \int\limits_0^1 {{2^{{x^2}}}} dx,\,{I_2} = \int\limits_0^1 {{2^{{x^3}}}} dx,$$
<br><br>$$ = {I_3} = \int\limits_0^1 {{2^{{x^2}}}} dx,\,$$
<br><br>$${I_4} = \int\limits_0^1 {{2^{{x^3}}}} dx\,\,$$
<br><br>$$\forall 0 < x < 1,\,{x^2} > {x^3}$$
<br><br>$$ \Rightarrow \int\limits_0^1 {{2^{{x^2}}}} \,dx > \int\limits_0^1 {{2^{{x^3}}}} dx$$
<br><br>and $$\int\limits_1^2 {{2^{{x^3}}}dx} > \int\limits_1^2 {{2^{{x^2}}}dx} $$
<br><br>$$ \Rightarrow {I_1} > {I_2}$$ and $${I_4} > {I_3}$$ | mcq | aieee-2005 |
uTC1jo5ctqq7OaZR | maths | definite-integration | properties-of-definite-integration | The value of $$\int\limits_{ - \pi }^\pi {{{{{\cos }^2}} \over {1 + {a^x}}}dx,\,\,a > 0,} $$ is | [{"identifier": "A", "content": "$$a\\,\\pi $$ "}, {"identifier": "B", "content": "$${\\pi \\over 2}$$ "}, {"identifier": "C", "content": "$${\\pi \\over a}$$"}, {"identifier": "D", "content": "$${2\\pi }$$ "}] | ["B"] | null | Let $$I = \int\limits_{ - \pi }^\pi {{{{{\cos }^2}x} \over {1 + {a^x}}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>$$ = \int\limits_{ - \pi }^\pi {{{{{\cos }^2}\left( { - x} \right)} \over {1 + {a^{ - x}}}}} dx$$
<br><br>$$\left[ \, \right.$$ Using $$\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)dx} $$ $$\left. \, \right]$$
<br><br>$$ = \int\limits_{ - \pi }^\pi {{{{{\cos }^2}x} \over {1 + {\alpha ^x}}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
<br><br>Adding equations $$(1)$$ and $$(2)$$ we get
<br><br>$$2I = \int\limits_{ - \pi }^\pi {{{\cos }^2}} x\left( {{{1 + {a^x}} \over {1 + {a^x}}}} \right)dx$$
<br><br>$$ = \int\limits_{ - \pi }^\pi {{{\cos }^2}} x\,dx$$
<br><br>$$ = 2\int\limits_0^\pi {{{\cos }^2}} x\,dx$$
<br><br>$$ = 2 \times 2\int\limits_0^{{\pi \over 2}} {{{\cos }^2}} x\,dx$$
<br><br>$$ = 4\int\limits_0^{{\pi \over 2}} {{{\sin }^2}} x\,dx$$
<br><br>$$ \Rightarrow I = 2\int\limits_0^{{\pi \over 2}} {{{\sin }^2}} \,x\,dx$$
<br><br>$$ = 2\int\limits_0^{{\pi \over 2}} {\left( {1 - {{\cos }^2}x\,dx} \right)} $$
<br><br>$$ \Rightarrow I\,\, = 2\int\limits_0^{{\pi \over 2}} {dx - 2\int\limits_0^{{\pi \over 2}} {{{\cos }^2}} } \,x\,dx$$
<br><br>$$ \Rightarrow I + I = 2\left( {{\pi \over 2}} \right) = \pi $$
<br><br>$$ \Rightarrow I = {\pi \over 2}$$ | mcq | aieee-2005 |
hYawqjqSy59anVIp | maths | definite-integration | properties-of-definite-integration | The value of integral, $$\int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }}} dx $$ is | [{"identifier": "A", "content": "$${1 \\over 2}$$ "}, {"identifier": "B", "content": "$${3 \\over 2}$$"}, {"identifier": "C", "content": "$$2$$"}, {"identifier": "D", "content": "$$1$$"}] | ["B"] | null | $$I = \int\limits_3^6 {{{\sqrt x } \over {\sqrt {9 - x} + \sqrt x }}} dx\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>$$I = \int\limits_3^6 {{{\sqrt {9 - x} } \over {\sqrt {9 - x} + \sqrt x }}} dx\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
<br><br>$$\left[ \, \right.$$ using $$\int\limits_a^b {f\left( x \right)dx = \int\limits_a^b {f\left( {a + b - x} \right)dx} } $$ $$\left. \, \right]$$
<br><br>Adding equation $$(1)$$ and $$(2)$$
<br><br>$$2I = \int\limits_3^6 {dx} = 3 \Rightarrow I = {3 \over 2}$$ | mcq | aieee-2005 |
EDjmCbMZawF5OenO | maths | definite-integration | properties-of-definite-integration | $$\int\limits_0^\pi {xf\left( {\sin x} \right)dx} $$ is equal to | [{"identifier": "A", "content": "$$\\pi \\int\\limits_0^\\pi {f\\left( {\\cos x} \\right)dx} $$ "}, {"identifier": "B", "content": "$$\\,\\pi \\int\\limits_0^\\pi {f\\left( {sinx} \\right)dx} $$ "}, {"identifier": "C", "content": "$${\\pi \\over 2}\\int\\limits_0^{\\pi /2} {f\\left( {sinx} \\right)dx} $$ "}, {"identifier": "D", "content": "$$\\pi \\int\\limits_0^{\\pi /2} {f\\left( {\\cos x} \\right)dx} $$ "}] | ["D"] | null | $$I = \int\limits_0^\pi {xf\left( {\sin \,x} \right)dx} $$
<br><br>$$ = \int\limits_0^\pi {\left( {\pi - x} \right)f\left( {\sin x} \right)dx} $$
<br><br>$$ = \pi \int\limits_0^\pi {f\left( {\sin x} \right)dx - 1} $$
<br><br>$$ \Rightarrow 2I = \pi {\pi \over 0}f\left( {\sin x} \right)dx$$
<br><br>$$I = {\pi \over 2}\int\limits_0^\pi {f\left( {\sin x} \right)dx} $$
<br><br>$$ = \pi \int\limits_0^{\pi /2} {f\left( {\sin x} \right)dx} $$
<br><br>$$ = \pi \int\limits_0^{\pi /2} {f\left( {\cos x} \right)dx} $$ | mcq | aieee-2006 |
83nydnA4phzxG3p2 | maths | definite-integration | properties-of-definite-integration | $$\int\limits_{ - {{3\pi } \over 2}}^{ - {\pi \over 2}} {\left[ {{{\left( {x + \pi } \right)}^3} + {{\cos }^2}\left( {x + 3\pi } \right)} \right]} dx$$ is equal to | [{"identifier": "A", "content": "$${{{\\pi ^4}} \\over {32}}$$ "}, {"identifier": "B", "content": "$${{{\\pi ^4}} \\over {32}} + {\\pi \\over 2}$$ "}, {"identifier": "C", "content": "$${\\pi \\over 2}$$ "}, {"identifier": "D", "content": "$${\\pi \\over 4} - 1$$ "}] | ["C"] | null | $$I = \int\limits_{ - {{3\pi } \over 2}}^{ - {\pi \over 2}} {\left[ {{{\left( {x + \pi } \right)}^3} + {{\cos }^2}\left( {x + 3\pi } \right)} \right]} \,dx$$
<br><br>Put $$x + \pi = t$$
<br><br>$$I = \int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{t^3} + {{\cos }^2}t} \right)dt} $$
<br><br>$$ = 2\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {{{\cos }^2}} tdt$$
<br><br>$$\left[ {} \right.$$ using the property of even and odd function $$\left. {} \right]$$
<br><br>$$ = \int\limits_0^{{\pi \over 2}} {\left( {1 + \cos 2t} \right)} dt = {\pi \over 2} + 0$$ | mcq | aieee-2006 |
7ans7jSH92bUnBqo | maths | definite-integration | properties-of-definite-integration | The value of $$\int\limits_1^a {\left[ x \right]} f'\left( x \right)dx,a > 1$$ where $${\left[ x \right]}$$ denotes the greatest integer not exceeding $$x$$ is | [{"identifier": "A", "content": "$$af\\left( a \\right) - \\left\\{ {f\\left( 1 \\right) + f\\left( 2 \\right) + .............f\\left( {\\left[ a \\right]} \\right)} \\right\\}$$ "}, {"identifier": "B", "content": "$$\\left[ a \\right]f\\left( a \\right) - \\left\\{ {f\\left( 1 \\right) + f\\left( 2 \\right) + ...........f\\left( {\\left[ a \\right]} \\right)} \\right\\}$$ "}, {"identifier": "C", "content": "$$\\left[ a \\right]f\\left( {\\left[ a \\right]} \\right) - \\left\\{ {f\\left( 1 \\right) + f\\left( 2 \\right) + ...........f\\left( a \\right)} \\right\\}$$ "}, {"identifier": "D", "content": "$$af\\left( {\\left[ a \\right]} \\right) - \\left\\{ {f\\left( 1 \\right) + f\\left( 2 \\right) + .............f\\left( a \\right)} \\right\\}$$ "}] | ["B"] | null | Let $$a = k + h$$ where $$k$$ is an integer such that
<br><br>$$\left[ a \right] = k$$ and $$0 \le h < 1$$
<br><br>$$\therefore$$ $$\int\limits_1^a {\left[ x \right]f'\left( x \right)dx = \int\limits_1^2 {1f'\left( x \right)} } \,dx$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\, + \int\limits_2^3 {2f'\left( x \right)dx + } ....$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\int\limits_{k - 1}^k {\left( {k - 1} \right)dx + \int\limits_k^{k + h} {kf'\left( x \right)} } dx$$
<br><br>$$\left\{ {f\left( 2 \right) - f\left( 1 \right)} \right\} + 2\left\{ {f\left( 3 \right) - f\left( 2 \right)} \right\}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ + 3\left\{ {f\left( 4 \right) - f\left( 3 \right)} \right\} + \,........\,$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ + \left( {k - 1} \right)\left\{ {f\left( k \right) - f\left( {k - 1} \right)} \right\}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ + k\left\{ {f\left( {k + h} \right) - f\left( k \right)} \right\}$$
<br><br>$$ = - f\left( 1 \right) - f\left( 2 \right) - f\left( 3 \right).......$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ - f\left( k \right) + kf\left( {k + h} \right)$$
<br><br>$$ = \left[ a \right]f\left( a \right) - \left\{ {f\left( 1 \right) + f\left( 2 \right)} \right.$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\left. { + f\left( 3 \right) + ........f\left( {\left[ a \right]} \right)} \right\}$$ | mcq | aieee-2006 |
G92XWMOvhmMdnIWk | maths | definite-integration | properties-of-definite-integration | The solution for $$x$$ of the equation $$\int\limits_{\sqrt 2 }^x {{{dt} \over {t\sqrt {{t^2} - 1} }} = {\pi \over 2}} $$ is | [{"identifier": "A", "content": "$${{\\sqrt 3 } \\over 2}$$ "}, {"identifier": "B", "content": "$$2\\sqrt 2 $$ "}, {"identifier": "C", "content": "$$2$$ "}, {"identifier": "D", "content": "None "}] | ["D"] | null | $$\int_{\sqrt 2 }^x {{{dt} \over {t\sqrt {{t^2} - 1} }}} = {\pi \over 2}$$
<br><br>$$\therefore$$ $$\left[ {{{\sec }^{ - 1}}t} \right]_{\sqrt 2 }^x = {\pi \over 2}$$
<br><br>$$\left[ {} \right.$$ As $$\int {{{dx} \over {x\sqrt {{x^2} - 1} }}} = {\sec ^{ - 1}}x$$ $$\left. {} \right]$$
<br><br>$$ \Rightarrow {\sec ^{ - 1}}x - {\sec ^{ - 1}}\sqrt 2 = {\pi \over 2}$$
<br><br>$$ \Rightarrow {\sec ^{ - 1}}x - {\pi \over 4} = {\pi \over 2}$$
<br><br>$$ \Rightarrow {\sec ^{ - 1}}x = {\pi \over 2} + {\pi \over 4}$$
<br><br>$$ \Rightarrow {\sec ^{ - 1}}x = {{3\pi } \over 4}$$
<br><br>$$ \Rightarrow x = \sec {{3\pi } \over 4}$$
<br><br>$$ \Rightarrow x = - \sqrt 2 $$ | mcq | aieee-2007 |
mzirftcb0gLb9BvX | maths | definite-integration | properties-of-definite-integration | Let $$I = \int\limits_0^1 {{{\sin x} \over {\sqrt x }}dx} $$ and $$J = \int\limits_0^1 {{{\cos x} \over {\sqrt x }}dx} .$$ Then which one of the following is true? | [{"identifier": "A", "content": "$$1 > {2 \\over 3}$$ and $$J > 2$$ "}, {"identifier": "B", "content": "$$1 < {2 \\over 3}$$ and $$J < 2$$ "}, {"identifier": "C", "content": "$$1 < {2 \\over 3}$$ and $$J > 2$$ "}, {"identifier": "D", "content": "$$1 > {2 \\over 3}$$ and $$J < 2$$ "}] | ["B"] | null | We know that $${{\sin x} \over x} < 1,$$ for $$x \in \left( {0,1} \right)$$
<br><br>$$ \Rightarrow {{\sin x} \over {\sqrt x }} < \sqrt x $$ on $$x \in \left( {0,1} \right)$$
<br><br>$$ \Rightarrow \int\limits_0^1 {{{\sin x} \over {\sqrt x }}dx < \int\limits_0^1 {\sqrt x dx} = \left[ {{{2{x^{3/2}}} \over 3}} \right]} _0^1$$
<br><br>$$ \Rightarrow \int\limits_0^1 {{{\sin x} \over {\sqrt x }}} dx < {2 \over 3} \Rightarrow I < {2 \over 3}$$
<br><br>Also $${{\cos x} \over {\sqrt x }} < {1 \over {\sqrt x }}$$ for $$x \in \left( {0,1} \right)$$
<br><br>$$ \Rightarrow \int\limits_0^1 {{{\cos x} \over {\sqrt x }}dx < \int\limits_0^1 {{x^{ - 1/2}}dx} } $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {2\sqrt x } \right]_0^1 = 2$$
<br><br>$$ \Rightarrow \int\limits_0^1 {{{\cos x} \over {\sqrt x }}dx < 2} $$
<br><br>$$ \Rightarrow J < 2$$ | mcq | aieee-2007 |
T2DpS7dxrPgx0DCQ | maths | definite-integration | properties-of-definite-integration | Let $$F\left( x \right) = f\left( x \right) + f\left( {{1 \over x}} \right),$$ where $$f\left( x \right) = \int\limits_l^x {{{\log t} \over {1 + t}}dt,} $$ Then $$F(e)$$ equals | [{"identifier": "A", "content": "$$1$$"}, {"identifier": "B", "content": "$$2$$"}, {"identifier": "C", "content": "$$1/2$$ "}, {"identifier": "D", "content": "$$0$$"}] | ["C"] | null | Given $$f\left( x \right) = f\left( x \right) + f\left( {{1 \over x}} \right),$$
<br><br>where $$f\left( x \right) = \int_1^x {{{\log \,t} \over {1 + t}}} \,dt$$
<br><br>$$\therefore$$ $$F\left( e \right) = f\left( e \right) + f\left( {{1 \over e}} \right)$$
<br><br>$$ \Rightarrow F\left( e \right)$$
<br><br>$$ = \int_1^e {{{\log \,t} \over {1 + t}}dt + \int_1^{1/e} {{{\log t} \over {1 + t}}} } dt\,\,\,....\left( A \right)$$
<br><br>Now for solving, $$I = \int_1^{1/e} {{{\log t} \over {1 + t}}dt} $$
<br><br>$$\therefore$$ Put $${1 \over t} = z \Rightarrow - {1 \over {{t^2}}}dt = dz$$
<br><br>$$ \Rightarrow dt = - {{dz} \over {{z^2}}}$$
<br><br>and limit for $$t = 1 \Rightarrow z = 1$$ and for
<br><br>$$t = 1/e \Rightarrow z = e$$
<br><br>$$\therefore$$ $$I = \int_1^e {{{\log \left( {{1 \over z}} \right)} \over {1 + {1 \over z}}}} \left( { - {{dz} \over {{z^2}}}} \right)$$
<br><br>$$ = \int_1^e {{{\left( {\log 1 - \log z} \right).z} \over {z + 1}}\left( { - {{dz} \over {{z^2}}}} \right)} $$
<br><br>$$ = \int_1^e { - {{\log z} \over {\left( {z + 1} \right)}}\left( { - {{dz} \over z}} \right)} $$
<br><br>[ as $$\log 1 = 0$$ ]
<br><br>$$ = \int_1^e {{{\log z} \over {z\left( {z + 1} \right)}}} dz$$
<br><br>$$\therefore$$ $$I = \int_1^e {{{\log \,t} \over {t\left( {t + 1} \right)}}dt} $$
<br><br>$$\left[ {} \right.$$ By property $$\int_a^b {f\left( t \right)dt} $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = \int_a^b {f\left( x \right)dx} $$ $$\left. {} \right]$$
<br><br>Equation $$(A)$$ becomes
<br><br>$$F\left( e \right) = \int_1^e {{{\log t} \over {1 + t}}dt + \int_1^e {{{\log t} \over {t\left( {1 + t} \right)}}} } dt$$
<br><br>$$ = \int_1^e {{{t.\log t + \log t} \over {t\left( {1 + t} \right)}}} dt$$
<br><br>$$ = \int_1^e {{{\left( {\log t} \right)\left( {t + 1} \right)} \over {t\left( {1 + t} \right)}}} $$
<br><br>$$ \Rightarrow F\left( e \right) = \int_1^e {{{\log t} \over t}dt} $$
<br><br>Let $$\log t = x$$
<br><br>$$\therefore$$ $${1 \over t}dt = dx$$
<br><br>$$\left[ {} \right.$$ for limit $$t = 1,x = 0$$
<br><br>and $$t = e,x = \log \,e = 1$$ $$\left. {} \right]$$
<br><br>$$\therefore$$ $$F\left( e \right) = \int_0^1 {x\,dx} $$
<br><br>$$ \Rightarrow F\left( e \right) = \left[ {{{{x^2}} \over 2}} \right]_0^1$$
<br><br>$$ \Rightarrow F\left( e \right) = {1 \over 2}$$ | mcq | aieee-2007 |
qzGZ9MQkHcseTyc4 | maths | definite-integration | properties-of-definite-integration | $$\int\limits_0^\pi {\left[ {\cot x} \right]dx,} $$ where $$\left[ . \right]$$ denotes the greatest integer function, is equal to: | [{"identifier": "A", "content": "$$1$$"}, {"identifier": "B", "content": "$$-1$$ "}, {"identifier": "C", "content": "$$ - {\\pi \\over 2}$$ "}, {"identifier": "D", "content": "$$ {\\pi \\over 2}$$"}] | ["C"] | null | Let $$I = \int_0^\pi {\left[ {\cot x} \right]dx\,\,\,\,\,\,...\left( 1 \right)} $$
<br><br>$$ = \int_0^\pi {\left[ {\cot \left( {\pi - x} \right)} \right]} dx$$
<br><br>$$ = \int_0^\pi {\left[ { - \cot x} \right]dx\,\,\,\,\,\,...\left( 2 \right)} $$
<br><br>Adding two values of $$I$$ in $$e{q^n}s\left( 1 \right)\,\,\& \,\,\left( 2 \right),$$
<br><br>We get $$2I = \int_0^\pi {\left( {\left[ {\cot x} \right] + \left[ { - \cot x} \right]} \right)dx} $$
<br><br>$$ = \int_0^\pi {\left( { - 1} \right)dx} $$
<br><br>$$\left[ {} \right.$$ As $$\left[ x \right] + \left[ { - x} \right] = - 1,\,\,if\,\,$$ $$x \notin z$$
<br><br>and $$\left[ x \right] + \left[ { - x} \right] = 0,\,\,if\,\,x \in z$$ $$\left. {} \right]$$
<br><br>$$ = \left[ { - x} \right]_0^\pi = - \pi \Rightarrow I = - {\pi \over 2}$$ | mcq | aieee-2009 |
vm6dQ1l0S3JOzXyO | maths | definite-integration | properties-of-definite-integration | Let $$p(x)$$ be a function defined on $$R$$ such that $$p'(x)=p'(1-x),$$ for all $$x \in \left[ {0,1} \right],p\left( 0 \right) = 1$$ and $$p(1)=41.$$ Then $$\int\limits_0^1 {p\left( x \right)dx} $$ equals : | [{"identifier": "A", "content": "$$21$$"}, {"identifier": "B", "content": "$$41$$ "}, {"identifier": "C", "content": "$$42$$ "}, {"identifier": "D", "content": "$$\\sqrt {41} $$ "}] | ["A"] | null | $$p'\left( x \right) = p'\left( {1 - x} \right)$$
<br><br>$$ \Rightarrow p\left( x \right) = - p\left( {1 - x} \right) + c$$
<br><br>at $$x=0$$
<br><br>$$p\left( 0 \right) = - p\left( 1 \right) + c \Rightarrow 42 = c$$
<br><br>Now, $$p\left( x \right) = - p\left( {1 - x} \right) + 42$$
<br><br>$$ \Rightarrow p\left( x \right) + p\left( {1 - x} \right) = 42$$
<br><br>$$ \Rightarrow I = \int\limits_0^1 {p\left( x \right)dx\,\,\,\,\,\,\,\,\,\,...\left( i \right)} $$
<br><br>$$ \Rightarrow I = \int\limits_0^1 {p\left( {1 - x} \right)} dx\,\,\,\,\,\,...\left( {ii} \right)$$
<br><br>on adding $$(i)$$ and $$(ii),$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,$$$$2I = \int\limits_0^1 {\left( {42} \right)dx \Rightarrow I = 21} $$ | mcq | aieee-2010 |
vJfDoRzgVdTiREgH | maths | definite-integration | properties-of-definite-integration | The value of $$\int\limits_0^1 {{{8\log \left( {1 + x} \right)} \over {1 + {x^2}}}} dx$$ is | [{"identifier": "A", "content": "$${\\pi \\over 8}\\log 2$$ "}, {"identifier": "B", "content": "$${\\pi \\over 2}\\log 2$$"}, {"identifier": "C", "content": "$$\\log 2$$ "}, {"identifier": "D", "content": "$$\\pi \\log 2$$ "}] | ["D"] | null | $$I = \int\limits_0^1 {{{8\log \left( {1 + x} \right)} \over {1 + {x^2}}}} dx$$
<br><br>put $$x = \tan \,\theta ,$$
<br><br>$$\therefore$$ $${{dx} \over {d\theta }} = {\sec ^2}\theta \Rightarrow dx = {\sec ^2}\theta d\theta $$
<br><br>$$\therefore$$ $$I = 8\int\limits_0^{\pi /4} {{{\log \left( {1 + \tan \theta } \right)} \over {1 + {{\tan }^2}\theta }}} .{\sec ^2}\theta d\theta $$
<br><br>$$I = 8\int\limits_0^{\pi /4} {\log \left( {1 + \tan \theta } \right)} d\theta \,...\left( i \right)$$
<br><br>$$ = 8\int\limits_0^{\pi /4} {\log \left[ {1 + \tan \left( {{\pi \over 4} - \theta } \right)} \right]} d\theta $$
<br><br>$$ = 8\int\limits_0^{\pi /4} {\log \left[ {1 + {{1 - \tan \theta } \over {1 + \tan \theta }}} \right]} d\theta $$
<br><br>$$ = 8\int\limits_0^{\pi /4} {\log \left[ {{2 \over {1 + \tan \theta }}} \right]} d\theta $$
<br><br>$$ = 8\int\limits_0^{\pi /4} {\left[ {\log 2 - \log \left( {1 + \tan \theta } \right)} \right]d\theta } $$
<br><br>$$I = 8.\left( {\log 2} \right)\left[ x \right]_0^{\pi /4} - 8$$
<br><br>$$\,\,\,\,\,\,\,\,\,\int\limits_0^{\pi /4} {\log \left( {1 + \tan \theta } \right)} d\theta $$
<br><br>$$I = 8.{\pi \over 4}.\log 2 - I$$ $$\left[ {} \right.$$ From equation $$(i)$$ $$\left. {} \right]$$
<br><br>$$ \Rightarrow 2I = 2\pi \log 2,$$
<br><br>$$\therefore$$ $$I = \pi \log 2$$ | mcq | aieee-2011 |
zUWPh9e7UZjwhvpd | maths | definite-integration | properties-of-definite-integration | If $$g\left( x \right) = \int\limits_0^x {\cos 4t\,dt,} $$ then $$g\left( {x + \pi } \right)$$ equals | [{"identifier": "A", "content": "$${{g\\left( x \\right)} \\over {8\\left( \\pi \\right)}}$$ "}, {"identifier": "B", "content": "$$g\\left( x \\right) + g\\left( \\pi \\right)$$ "}, {"identifier": "C", "content": "$$g\\left( x \\right) - g\\left( \\pi \\right)$$"}, {"identifier": "D", "content": "$$g\\left( x \\right) . g\\left( \\pi \\right)$$"}] | null | null | $$\left( {b,c} \right)g\left( {x + \pi } \right) = \int\limits_0^{x + \pi } {\cos \,4t\,dt} $$
<br><br>$$ = \int\limits_0^\pi {\cos 4tdt + \int\limits_\pi ^{\pi + x} {\cos 4t\,dt} } $$
<br><br>$$ = g\left( \pi \right) + \int\limits_0^x {\cos \,4t\,dt} $$
<br><br>Putting $$t = \pi + y$$ in second integral, we get
<br><br>$$\int\limits_x^{\pi + x} {\cos \,4t\,dt = \int\limits_0^\pi {\cos \,4t\,dt} } $$
<br><br>$$ = g\left( \pi \right) + g\left( x \right) = g\left( x \right) - g\left( \pi \right)$$
<br><br>$$\therefore$$ $$\left. {g\left( \pi \right) = 0} \right)$$ | mcqm | aieee-2012 |
ntDZ2U0Brhjp26Vh | maths | definite-integration | properties-of-definite-integration | <b>Statement-1 :</b> The value of the integral
<br/>$$\int\limits_{\pi /6}^{\pi /3} {{{dx} \over {1 + \sqrt {\tan \,x} }}} $$ is equal to $$\pi /6$$
<p><b>Statement-2 :</b> $$\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx.$$</p> | [{"identifier": "A", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1."}, {"identifier": "B", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1."}, {"identifier": "C", "content": "Statement- 1 is true; Statement-2 is False."}, {"identifier": "D", "content": "Statement-1 is false; Statement-2 is true."}] | ["D"] | null | Let $$I = \int\limits_{\pi /6}^{\pi /3} {{{dx} \over {1 + \sqrt {\tan \,x} }}} $$
<br><br>$$ = \int\limits_{\pi /6}^{\pi /3} {{{dx} \over {\sqrt {\tan \left( {{\pi \over 2} - x} \right)} }}} $$
<br><br>$$ = \int\limits_{\pi /6}^{\pi /3} {{{\sqrt {\tan \,x} \,dx} \over {1 + \sqrt {\tan \,x} }}} \,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>Also, Given,
<br><br>$$I$$ $$ = \int\limits_{\pi /6}^{\pi /3} {{{\sqrt {\tan \,x} \,dx} \over {1 + \sqrt {\tan \,x} }}} \,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
<br><br>By adding $$(1)$$ and $$(2),$$ we get
<br><br>$$2I = \int\limits_{\pi /6}^{\pi /3} {dx} $$
<br><br>$$ \Rightarrow I = {1 \over 2}\left[ {{\pi \over 3} - {\pi \over 6}} \right]$$
<br><br>$$ = {\pi \over {12}},$$ statements -$$1$$ is false
<br><br>$$\int\limits_a^b {f\left( x \right)dx = \int\limits_a^b {f\left( {a + b - x} \right)dx} } $$
<br><br>It is fundamental property. | mcq | jee-main-2013-offline |
4BKFMfmAUvUVke3R | maths | definite-integration | properties-of-definite-integration | The integral $$\int\limits_0^\pi {\sqrt {1 + 4{{\sin }^2}{x \over 2} - 4\sin {x \over 2}{\mkern 1mu} } } dx$$ equals: | [{"identifier": "A", "content": "$$4\\sqrt 3 - 4$$ "}, {"identifier": "B", "content": "$$4\\sqrt 3 - 4 - {\\pi \\over 3}$$ "}, {"identifier": "C", "content": "$$\\pi - 4$$ "}, {"identifier": "D", "content": "$${{2\\pi } \\over 3} - 4 - 4\\sqrt 3 $$ "}] | ["B"] | null | Let $$I = \int\limits_0^\pi {\sqrt {1 + 4{{\sin }^2}{x \over 2} - 4\sin {x \over 2}} } dx$$
<br><br>$$ = \int\limits_0^\pi {\left| {2\sin {x \over 2} - 1} \right|} dx$$
<br><br>$$ = \int\limits_0^{\pi /3} {\left( {1 - 2\sin {x \over 2}} \right)} dx + \int\limits_{\pi /3}^\pi {\left( {2\sin {x \over 2} - 1} \right)} dx$$
<br><br>$$\left[ {} \right.$$ As $$\sin {x \over 2} = {1 \over 2} \Rightarrow {x \over 2} = {\pi \over 6} \Rightarrow x = {\pi \over 3},$$ $${x \over 2}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\, = {{5\pi } \over 6} \Rightarrow x = {{5\pi } \over 3}$$ $$\left. {} \right]$$
<br><br>$$ = \left[ {x + 4\cos {x \over 2}} \right]_0^{\pi /3} + \left[ { - 4\cos {x \over 2} - x} \right]_{\pi /3}^\pi $$
<br><br>$$ = {\pi \over 3} + 4{{\sqrt 2 } \over 2} - 4 + \left( {0 - \pi + 4{{\sqrt 3 } \over 2} + {\pi \over 3}} \right)$$
<br><br>$$ = 4\sqrt 3 - 4 - {\pi \over 3}$$ | mcq | jee-main-2014-offline |
t8g0J8Q5j5Uqh3wN | maths | definite-integration | properties-of-definite-integration | The integral
<br/>$$\int\limits_2^4 {{{\log \,{x^2}} \over {\log {x^2} + \log \left( {36 - 12x + {x^2}} \right)}}dx} $$ is equal to : | [{"identifier": "A", "content": "$$1$$"}, {"identifier": "B", "content": "$$6$$"}, {"identifier": "C", "content": "$$2$$"}, {"identifier": "D", "content": "$$4$$"}] | ["A"] | null | $$I = \int\limits_2^4 {{{\log {x^2}} \over {\log {x^2} + \log \left( {36 - 12x + {x^2}} \right)}}} $$
<br><br>$$I = \int\limits_2^4 {{{\log {x^2}} \over {\log {x^2} + \log {{\left( {6 - x} \right)}^2}}}} \,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>$$I = \int\limits_2^4 {{{\log {{\left( {6 - x} \right)}^2}} \over {\log {{\left( {6 - x} \right)}^2} + \log {x^2}}}} \,\,\,\,\,\,\,\,...\left( {ii} \right)$$
<br><br>Adding $$(1)$$ and $$(2)$$
<br><br>$$2I = \int\limits_2^4 {dx = \left[ x \right]_2^4} = 2 \Rightarrow 1 - 1$$ | mcq | jee-main-2015-offline |
m9JhdOqShTaju1uZ4jGNd | maths | definite-integration | properties-of-definite-integration | The value of the integral
<br/><br/>$$\int\limits_4^{10} {{{\left[ {{x^2}} \right]dx} \over {\left[ {{x^2} - 28x + 196} \right] + \left[ {{x^2}} \right]}}} ,$$
<br/><br/>where [x] denotes the greatest integer less than or
equal to x, is : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "$${1 \\over 3}$$ "}] | ["B"] | null | Let I = $$\int\limits_4^{10} {{{\left[ {{x^2}} \right]\,dx} \over {\left[ {{x^2} - 28x + 196} \right] + \left[ {{x^2}} \right]}}} $$
<br><br> = $$\int\limits_4^{10} {{{\left[ {{x^2}} \right]dx} \over {\left[ {{{\left( {x - 14} \right)}^2}} \right] + \left[ {{x^2}} \right]}}} \,\,\,.....(1)$$
<br><br>Using,
<br><br>$$\int\limits_a^b {f\left( {a + b - x} \right)dx\,} $$ = $$\,\,\int\limits_a^b {f(x)\,\,dx} $$
<br><br>I = $$\int\limits_4^{10} {{{{{\left( {x - 14} \right)}^2}} \over {\left[ {{x^2}} \right] + \left[ {{{\left( {x - 14} \right)}^2}} \right]}}} \,dx\,\,....(2)$$
<br><br>Adding (1) and (2)
<br><br>2I = $$\int\limits_4^{10} {{{\left[ {{{\left( {x - 14} \right)}^2}} \right] + \left[ {{x^2}} \right]} \over {\left[ {{x^2}} \right] + \left[ {{{\left( {x - 14} \right)}^2}} \right]}}} \,dx$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 2I = $$\int\limits_4^{10} {dx} = \left[ x \right]_4^{10}$$ = 6
<br><br>= I = 3 | mcq | jee-main-2016-online-10th-april-morning-slot |
5PeyGagsIrrEEA35wifL0 | maths | definite-integration | properties-of-definite-integration | If $$2\int\limits_0^1 {{{\tan }^{ - 1}}xdx = \int\limits_0^1 {{{\cot }^{ - 1}}} } \left( {1 - x + {x^2}} \right)dx,$$
<br/><br/>then $$\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx$$ is equalto : | [{"identifier": "A", "content": "log4"}, {"identifier": "B", "content": "$${\\pi \\over 2}$$ + log2"}, {"identifier": "C", "content": "log2"}, {"identifier": "D", "content": "$${\\pi \\over 2}$$ $$-$$ log4"}] | ["C"] | null | Given,
<br><br>$$2\int\limits_0^1 {{{\tan }^{ - {1_x}\,{d_x}}}} = \int\limits_0^1 {{{\cot }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx$$
<br><br>= $$\int\limits_0^1 {\left( {{\pi \over 2} - {{\tan }^{ - 1}}\left( {1 - x + {x^2}} \right)} \right)} dx$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$$$2\int\limits_0^1 {{{\tan }^{ - 1}}} x\,dx = \int\limits_0^1 {{\pi \over 2}} dx - \int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $$\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx = \int\limits_0^1 {{\pi \over 2}} dx - 2\int\limits_0^1 {{{\tan }^{ - 1}}x\,dx} $$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $$\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)dx = {\pi \over 2} - 2\int\limits_0^1 {{{\tan }^{ - 1}}x\,\,dx} $$
<br><br>Let I = $$\int\limits_0^1 {{{\tan }^{ - 1}}} x\,dx$$
<br><br>= $$\left[ {\left( {{{\tan }^{ - 1}}x} \right)x} \right]_0^1 - \int\limits_0^1 {{1 \over {1 + {x^2}}}} x\,dx$$
<br><br>= $${\pi \over 4} - {1 \over {20}}\left[ {\log \left| {1 + {x^2}} \right|} \right]_0^1$$
<br><br>= $${\pi \over 4} - {1 \over 2}\log 2$$
<br><br>$$\therefore\,\,\,$$ $$\int\limits_0^1 {{{\tan }^{ - 1}}} \left( {1 - x + {x^2}} \right)\,dx$$
<br><br>= $${\pi \over 2} - 2\left( {{\pi \over 4} - {1 \over 2}\log 2} \right)$$
<br><br>= $${\pi \over 2} - {\pi \over 2} + \log 2$$
<br><br>= log2 | mcq | jee-main-2016-online-9th-april-morning-slot |
HG6YBLywBgnrD491MxStk | maths | definite-integration | properties-of-definite-integration | The integral $$\int_{{\pi \over {12}}}^{{\pi \over 4}} {\,\,{{8\cos 2x} \over {{{\left( {\tan x + \cot x} \right)}^3}}}} \,dx$$ equals : | [{"identifier": "A", "content": "$${{15} \\over {128}}$$"}, {"identifier": "B", "content": "$${{15} \\over {64}}$$"}, {"identifier": "C", "content": "$${{13} \\over {32}}$$"}, {"identifier": "D", "content": "$${{13} \\over {256}}$$"}] | ["A"] | null | tan x + cot x
<br><br>= $${{\sin x} \over {\cos x}}$$ + $${{\cos x} \over {\sin x}}$$
<br><br>= $${{{{\sin }^2}x + {{\cos }^2}x} \over {\sin x\,\,\cos x}}$$
<br><br>= $${1 \over {\sin x\,\,\cos x}}$$
<br><br>= $${2 \over {\sin 2x}}$$
<br><br>$$\therefore\,\,\,$$ $$\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{8\cos 2x} \over {{{\left( {\tan x + \cot x} \right)}^3}}}} \,\,dx$$
<br><br>= $$\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{8\cos 2x} \over {{{\left( {{2 \over {\sin 2x}}} \right)}^3}}}} \,\,dx$$
<br><br>= $$\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{8{{\sin }^3}\,2x.\cos \,\,2x} \over 8}} \,\,dx$$
<br><br>= $$\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{\sin }^3}2x.\cos \,2x} \,\,dx$$
<br><br>Let sin 2x = t
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 2 cos2x dx = dt
<br><br>At x = $${{\pi \over {12}}}$$, t = sin $${{\pi \over {6}}}$$ = $${1 \over 2}$$
<br><br>At x = $${{\pi \over 4}}$$, t = sin $${{\pi \over 2}}$$ = 1.
<br><br>= $${1 \over 2}$$ $$\int\limits_{{1 \over 2}}^1 {{t^3}.dt} $$
<br><br>= $${{1 \over 2}}$$ $$\left[ {{{{t^4}} \over 4}} \right]_{{1 \over 2}}^1$$
<br><br>= $${{1 \over 2}}$$ $$ \times $$ $${{1 \over 4}}$$ $$ \times $$ [ 1<sup>4</sup> $$-$$ ($${{1 \over 2}}$$)<sup>4</sup>]
<br><br>= $${{1 \over 8}}$$ $$ \times $$ $${{15 \over 16}}$$ = $${{15 \over 128}}$$ | mcq | jee-main-2017-online-8th-april-morning-slot |
43iXG9zjRuTrzgRtZqrQT | maths | definite-integration | properties-of-definite-integration | If $$\int\limits_1^2 {{{dx} \over {{{\left( {{x^2} - 2x + 4} \right)}^{{3 \over 2}}}}}} = {k \over {k + 5}},$$ then k is equal to : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}] | ["A"] | null | Given, I = $$\int\limits_1^2 {{{dx} \over {{{\left( {{x^2} - 2x + 4} \right)}^{{3 \over 2}}}}}} $$
<br><br>= $$\int\limits_1^2 {{{dx} \over {{{\left[ {{{\left( {x - 1} \right)}^2} + 3} \right]}^{{3 \over 2}}}}}} $$
<br><br>Let x $$-$$ 1 = $$\sqrt 3 $$ tan$$\theta $$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ dx = $$\sqrt 3 $$ sec<sup>2</sup>$$\theta $$ d$$\theta $$
<br><br>When x = 1, then $$\theta $$ = 0
<br><br>and when x = 2, $$\theta $$ = $${\pi \over 6}$$
<br><br>I = $$\int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {{{\left( {3{{\tan }^2}\theta + 3} \right)}^{{3 \over 2}}}}}} $$
<br><br>= $$\int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {{{\left( {3{{\sec }^2}\theta } \right)}^{{3 \over 2}}}}}} $$
<br><br>= $$\int\limits_0^{{\pi \over 6}} {{{\sqrt 3 {{\sec }^2}\theta \,d\theta } \over {3\sqrt 3 {{\sec }^3}\theta }}} $$
<br><br>= $${1 \over 3}$$ $$\int\limits_0^{{\pi \over 6}} {{{d\theta } \over {\sec \theta }}} $$
<br><br>= $${1 \over 3}\int\limits_0^{{\pi \over 6}} {\cos \theta \,d\theta } $$
<br><br>= $${1 \over 3}{\left[ {\sin \theta } \right]_\theta }^{{\pi \over 6}}$$
<br><br>= $${1 \over 3} \times {1 \over 2}$$
<br><br>= $${1 \over 6}$$
<br><br>$$\therefore\,\,\,$$ According to the question,
<br><br>$${k \over {k + 5}}$$ = $${1 \over 6}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 6k = k + 5
<br><br>$$ \Rightarrow $$$$\,\,\,$$ k = 1 | mcq | jee-main-2017-online-9th-april-morning-slot |
q5elwpzUgF4Mve5d | maths | definite-integration | properties-of-definite-integration | The integral $$\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {1 + \cos x}}} $$ is equal to | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "$$-$$ 1"}, {"identifier": "D", "content": "$$-$$ 2"}] | ["A"] | null | $$\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {1 + \cos x}}} $$
<br><br>= $$\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {2{{\cos }^2}{x \over 2}}}} $$
<br><br>= $${1 \over 2}\int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{\sec }^2}} {x \over 2}\,dx$$
<br><br>$${1 \over 2}\left[ {{{\tan {x \over 2}} \over {{1 \over 2}}}} \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}$$
<br><br>= $$\left[ {\tan {x \over 2}} \right]_{{\pi \over 4}}^{{3 \over 4}}$$
<br><br>= tan $${{3\pi } \over 8}$$ $$-$$ tan$${\pi \over 8}$$
<br><br>= $$\left( {\sqrt 2 + 1} \right) - \left( {\sqrt 2 - 1} \right)$$
<br><br>= 2 | mcq | jee-main-2017-offline |
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